Q. A 10 kg block is sliding on a surface with a coefficient of kinetic friction of 0.3. What is the frictional force acting on the block?
A.
30 N
B.
15 N
C.
3 N
D.
0 N
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Solution
The frictional force can be calculated using F_friction = μ * N, where N = mg = 10 kg * 9.81 m/s² = 98.1 N. Thus, F_friction = 0.3 * 98.1 N = 29.43 N, approximately 30 N.
Correct Answer:
A
— 30 N
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Q. A 2 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.
9 J
B.
6 J
C.
3 J
D.
12 J
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Solution
Kinetic energy is given by KE = 0.5 * m * v² = 0.5 * 2 kg * (3 m/s)² = 9 J.
Correct Answer:
A
— 9 J
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Q. A ball is thrown vertically upwards with a speed of 20 m/s. How high will it go before it starts to fall back down? (g = 10 m/s²)
A.
20 m
B.
30 m
C.
40 m
D.
10 m
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Solution
Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -10 m/s², we get 0 = 20² + 2 * (-10) * s, solving gives s = 20 m.
Correct Answer:
B
— 30 m
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Q. A car accelerates from rest at a rate of 2 m/s². How far does it travel in 5 seconds?
A.
25 m
B.
20 m
C.
10 m
D.
15 m
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Solution
Using the formula s = ut + 0.5at², where u = 0, a = 2 m/s², and t = 5 s, we get s = 0 + 0.5 * 2 * 5² = 25 m.
Correct Answer:
A
— 25 m
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Q. A car moving at 60 km/h applies brakes and comes to a stop in 5 seconds. What is the distance covered during braking?
A.
50 m
B.
60 m
C.
70 m
D.
40 m
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Solution
Convert speed to m/s: 60 km/h = 16.67 m/s. Using s = ut + 0.5at², where a = (0 - 16.67)/5 = -3.33 m/s², we find s = 16.67 * 5 + 0.5 * (-3.33) * 5² = 41.67 m.
Correct Answer:
A
— 50 m
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Q. A car travels 100 meters in 4 seconds. What is its average speed?
A.
25 m/s
B.
20 m/s
C.
15 m/s
D.
30 m/s
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Solution
Average speed = total distance / total time = 100 m / 4 s = 25 m/s.
Correct Answer:
B
— 20 m/s
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Q. A cyclist travels at a speed of 15 m/s. How long will it take to cover a distance of 300 meters?
A.
20 s
B.
15 s
C.
25 s
D.
10 s
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Solution
Using the formula time = distance/speed, we have time = 300 m / 15 m/s = 20 s.
Correct Answer:
A
— 20 s
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Q. A force of 50 N is applied at an angle of 30 degrees to the horizontal. What is the horizontal component of the force?
A.
25 N
B.
43.3 N
C.
50 N
D.
0 N
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Solution
The horizontal component is given by F_horizontal = F * cos(θ) = 50 N * cos(30°) = 50 N * (√3/2) ≈ 43.3 N.
Correct Answer:
B
— 43.3 N
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Q. A particle moves in a circular path of radius 5 m with a constant speed of 10 m/s. What is the centripetal acceleration?
A.
2 m/s²
B.
10 m/s²
C.
20 m/s²
D.
5 m/s²
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Solution
Centripetal acceleration is given by a_c = v² / r. Here, a_c = (10 m/s)² / 5 m = 20 m/s².
Correct Answer:
C
— 20 m/s²
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Q. A projectile is launched at an angle of 30 degrees with an initial speed of 40 m/s. What is the maximum height reached? (g = 9.8 m/s²)
A.
80 m
B.
60 m
C.
40 m
D.
20 m
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Solution
The vertical component of the velocity is 40 * sin(30) = 20 m/s. Using h = (v²)/(2g), we get h = (20²)/(2 * 9.8) ≈ 20.41 m.
Correct Answer:
B
— 60 m
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Q. A projectile is launched at an angle of 30 degrees with an initial velocity of 20 m/s. What is the horizontal component of its velocity?
A.
10 m/s
B.
17.32 m/s
C.
20 m/s
D.
5 m/s
Show solution
Solution
The horizontal component is given by v_x = v * cos(θ) = 20 * cos(30°) = 20 * (√3/2) ≈ 17.32 m/s.
Correct Answer:
B
— 17.32 m/s
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Q. A train moving at 72 km/h applies brakes and comes to a stop in 10 seconds. What is the deceleration?
A.
2 m/s²
B.
3 m/s²
C.
4 m/s²
D.
5 m/s²
Show solution
Solution
First convert speed to m/s: 72 km/h = 20 m/s. Using a = (v - u)/t, we have a = (0 - 20)/10 = -2 m/s², so deceleration = 2 m/s².
Correct Answer:
C
— 4 m/s²
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Q. An object is in free fall from a height of 80 m. How long will it take to reach the ground? (g = 9.8 m/s²)
A.
4 s
B.
5 s
C.
6 s
D.
3 s
Show solution
Solution
Using the formula s = ut + 0.5gt², where u = 0, s = 80 m, and g = 9.8 m/s², we solve 80 = 0 + 0.5 * 9.8 * t², giving t ≈ 4.04 s.
Correct Answer:
B
— 5 s
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Q. An object is thrown horizontally from a height of 45 m. How long does it take to hit the ground? (g = 9.8 m/s²)
A.
3 s
B.
4 s
C.
5 s
D.
2 s
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Solution
Using the formula s = 0.5gt², we have 45 = 0.5 * 9.8 * t², solving gives t ≈ 3.03 s.
Correct Answer:
B
— 4 s
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Q. An object is thrown vertically upward with an initial velocity of 20 m/s. How high will it rise before coming to a stop?
A.
20 m
B.
40 m
C.
10 m
D.
30 m
Show solution
Solution
Using the formula h = (v² - u²) / (2g), where v = 0, u = 20 m/s, and g = 9.81 m/s², we find h = (0 - (20)²) / (2 * -9.81) = 20.39 m, approximately 40 m.
Correct Answer:
B
— 40 m
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Q. If a ball is thrown vertically upward with an initial velocity of 15 m/s, how high will it go before coming to a stop?
A.
11.47 m
B.
15 m
C.
7.5 m
D.
20 m
Show solution
Solution
Using the formula h = v²/(2g), where v = 15 m/s and g = 9.81 m/s², we find h = (15)²/(2*9.81) ≈ 11.47 m.
Correct Answer:
A
— 11.47 m
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Q. If a car accelerates from rest at a rate of 2 m/s², what is its velocity after 5 seconds?
A.
10 m/s
B.
5 m/s
C.
15 m/s
D.
20 m/s
Show solution
Solution
Using the formula v = u + at, where u = 0, a = 2 m/s², and t = 5 s, we get v = 0 + (2)(5) = 10 m/s.
Correct Answer:
A
— 10 m/s
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Q. If a car accelerates from rest at a rate of 3 m/s², what is its velocity after 5 seconds?
A.
15 m/s
B.
10 m/s
C.
5 m/s
D.
20 m/s
Show solution
Solution
Using the formula v = u + at, where u = 0, a = 3 m/s², and t = 5 s, we get v = 0 + (3)(5) = 15 m/s.
Correct Answer:
A
— 15 m/s
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Q. If a car travels 100 meters in 5 seconds, what is its average speed?
A.
20 m/s
B.
25 m/s
C.
15 m/s
D.
30 m/s
Show solution
Solution
Average speed is calculated as total distance divided by total time: 100 m / 5 s = 20 m/s.
Correct Answer:
B
— 25 m/s
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Q. In a frictionless environment, if a 5 kg block is pushed with a force of 20 N, what will be its acceleration?
A.
4 m/s²
B.
2 m/s²
C.
5 m/s²
D.
0 m/s²
Show solution
Solution
Using F = ma, we have 20 N = 5 kg * a, thus a = 20 N / 5 kg = 4 m/s².
Correct Answer:
A
— 4 m/s²
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Q. What is the acceleration of an object in free fall near the Earth's surface?
A.
9.81 m/s²
B.
0 m/s²
C.
1.62 m/s²
D.
32.2 ft/s²
Show solution
Solution
The acceleration due to gravity near the Earth's surface is approximately 9.81 m/s².
Correct Answer:
A
— 9.81 m/s²
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Q. What is the center of gravity of a uniform beam of length L?
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Solution
For a uniform beam, the center of gravity is located at its midpoint, which is L/2.
Correct Answer:
B
— L/2
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Q. What is the formula for calculating the frictional force?
A.
F_friction = μN
B.
F_friction = m*a
C.
F_friction = N/g
D.
F_friction = v/t
Show solution
Solution
The frictional force is calculated using the formula F_friction = μN, where μ is the coefficient of friction and N is the normal force.
Correct Answer:
A
— F_friction = μN
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Q. What is the net force acting on a 10 kg object that is accelerating at 3 m/s²?
A.
30 N
B.
10 N
C.
3 N
D.
0 N
Show solution
Solution
Using Newton's second law, F = ma, we find F = 10 kg * 3 m/s² = 30 N.
Correct Answer:
A
— 30 N
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Q. What is the net force acting on a 5 kg object that is accelerating at 2 m/s²?
A.
10 N
B.
5 N
C.
2 N
D.
15 N
Show solution
Solution
Using Newton's second law, F = ma, where m = 5 kg and a = 2 m/s², we find F = 5 kg * 2 m/s² = 10 N.
Correct Answer:
A
— 10 N
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Q. What is the time taken for an object to fall from a height of 45 m under gravity?
A.
3 s
B.
4.5 s
C.
5 s
D.
6 s
Show solution
Solution
Using the formula h = 0.5 * g * t², we can rearrange to find t = √(2h/g). Substituting h = 45 m and g = 9.81 m/s² gives t ≈ 4.5 s.
Correct Answer:
B
— 4.5 s
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