Q. A beam of light passes through a thin convex lens with a focal length of 15 cm. If the object is placed 30 cm from the lens, what is the image distance?
A.10 cm
B.15 cm
C.20 cm
D.30 cm
Solution
Using the lens formula, 1/f = 1/v - 1/u; here, f = 15 cm and u = -30 cm. Thus, 1/v = 1/15 + 1/30 = 1/10, giving v = 10 cm.
Q. A block is at rest on a horizontal surface. If the applied force is gradually increased and reaches the maximum static frictional force, what will happen next?
A.The block will remain at rest
B.The block will start moving
C.The block will accelerate
D.The block will slide back
Solution
Once the applied force exceeds the maximum static frictional force, the block will start moving.
Q. A block is sliding down a frictionless incline of angle 30 degrees. If the incline has a coefficient of static friction of 0.5, what is the maximum angle at which the block can remain at rest?
A.30 degrees
B.45 degrees
C.60 degrees
D.90 degrees
Solution
The maximum angle for static friction is given by tan(θ) = μs. Here, θ = tan⁻¹(0.5) which is approximately 26.57 degrees, so the block can remain at rest at angles less than this.
Q. A block is sliding down a frictionless incline of angle θ. If the incline has a coefficient of static friction μs, what is the maximum angle θ for which the block will not slide?
A.tan⁻¹(μs)
B.sin⁻¹(μs)
C.cos⁻¹(μs)
D.μs
Solution
The block will not slide if the component of gravitational force down the incline is less than or equal to the maximum static friction force, leading to θ = tan⁻¹(μs).
Q. A block is sliding down a frictionless incline. If the incline is now covered with a material that has a coefficient of kinetic friction of 0.3, how does this affect the acceleration of the block?
A.Increases acceleration
B.Decreases acceleration
C.No effect on acceleration
D.Acceleration becomes zero
Solution
The presence of kinetic friction opposes the motion, thus decreasing the acceleration of the block compared to a frictionless incline.
Q. A block of mass 10 kg is resting on a horizontal surface. If the coefficient of kinetic friction is 0.3, what is the frictional force acting on the block when it is sliding?
A.30 N
B.20 N
C.10 N
D.15 N
Solution
Frictional force (f_k) = μ_k * N = μ_k * mg = 0.3 * 10 kg * 9.8 m/s² = 29.4 N, approximately 30 N.
Q. A block of mass 10 kg is resting on a horizontal surface. If the coefficient of static friction is 0.5, what is the maximum static frictional force acting on the block?
A.25 N
B.50 N
C.75 N
D.100 N
Solution
Maximum static frictional force (Fs) = μs * N = μs * mg = 0.5 * 10 kg * 9.8 m/s² = 49 N, approximately 50 N.
Q. A block of mass 2 kg is pushed along a frictionless surface by a constant force of 10 N. What is the work done by the force when the block moves 5 m?
Q. A block of mass 2 kg is pushed along a horizontal surface with a constant force of 10 N. What is the work done by the force after moving the block 5 m?
Q. A block of mass 2 kg is pushed along a horizontal surface with a force of 10 N. If the block moves a distance of 5 m, what is the work done by the force?
Q. A block of mass 2 kg is pushed along a horizontal surface with a force of 10 N. If the block moves a distance of 5 m, what is the work done on the block?
Q. A block of mass 2 kg is released from a height of 10 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.14 m/s
B.20 m/s
C.10 m/s
D.5 m/s
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 10) = 14 m/s.
Q. A block of mass 2 kg is released from a height of 10 m. What is its speed just before it hits the ground?
A.0 m/s
B.10 m/s
C.14 m/s
D.20 m/s
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*10) = 14 m/s.
Q. A block of mass 2 kg is released from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.5 m/s
B.10 m/s
C.15 m/s
D.20 m/s
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 5) = 10 m/s.
Q. A block of mass 5 kg is resting on a horizontal surface. If a horizontal force of 20 N is applied, what is the acceleration of the block? (Assume no friction)
A.2 m/s²
B.4 m/s²
C.5 m/s²
D.10 m/s²
Solution
Using Newton's second law, F = ma, we have a = F/m = 20 N / 5 kg = 4 m/s².
Q. A block of mass 5 kg is resting on a horizontal surface. If the coefficient of kinetic friction between the block and the surface is 0.3, what is the frictional force acting on the block when it is sliding?
A.5 N
B.10 N
C.15 N
D.20 N
Solution
Frictional force (Ff) = μk * N = μk * mg = 0.3 * (5 kg * 10 m/s²) = 15 N.
Q. A block of mass 5 kg is resting on a horizontal surface. If the coefficient of static friction is 0.4, what is the maximum static frictional force acting on the block?
A.10 N
B.20 N
C.15 N
D.25 N
Solution
Maximum static frictional force (Fs) = μs * N = μs * mg = 0.4 * (5 kg * 10 m/s²) = 20 N.
Q. A block on a frictionless surface is attached to a spring and undergoes simple harmonic motion. If the spring constant is 200 N/m and the mass is 2 kg, what is the period of oscillation?
A.0.5 s
B.1 s
C.2 s
D.4 s
Solution
The period T is given by T = 2π√(m/k). Here, T = 2π√(2/200) = 2π√(0.01) = 2π(0.1) = 0.2π ≈ 0.63 s.
Q. A block slides down a frictionless incline of angle 30 degrees. If the incline has a coefficient of kinetic friction of 0.2, what is the acceleration of the block?
A.4.9 m/s²
B.3.9 m/s²
C.2.9 m/s²
D.1.9 m/s²
Solution
Net force = mg sin(30) - μmg cos(30). Acceleration a = (mg sin(30) - μmg cos(30))/m = g(sin(30) - μ cos(30)). Substituting g = 10 m/s² gives a = 10(0.5 - 0.2 * √3/2) = 4.9 m/s².
Q. A boat can travel at 10 km/h in still water. If it is moving downstream in a river flowing at 5 km/h, what is the speed of the boat relative to the riverbank?
A.5 km/h
B.10 km/h
C.15 km/h
D.20 km/h
Solution
Speed downstream = Speed of boat + Speed of river = 10 km/h + 5 km/h = 15 km/h.
