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 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 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 box is pushed across a floor with a force of 50 N. If the coefficient of kinetic friction is 0.4, what is the net force acting on the box if the normal force is 100 N?
A.10 N
B.20 N
C.30 N
D.40 N
Solution
Frictional force = μk * N = 0.4 * 100 N = 40 N. Net force = applied force - frictional force = 50 N - 40 N = 10 N.
Q. A box is pushed with a force of 50 N on a surface with a coefficient of kinetic friction of 0.3. If the normal force is 100 N, what is the net force acting on the box?
A.20 N
B.30 N
C.50 N
D.70 N
Solution
Frictional force = μk * N = 0.3 * 100 N = 30 N. Net force = applied force - frictional force = 50 N - 30 N = 20 N.
Q. A box is pushed with a force of 50 N on a surface with a coefficient of kinetic friction of 0.4. What is the acceleration of the box if its mass is 10 kg?
A.1 m/s²
B.2 m/s²
C.3 m/s²
D.4 m/s²
Solution
Net force = applied force - frictional force. Frictional force = μ_k * N = 0.4 * 10 kg * 9.8 m/s² = 39.2 N. Net force = 50 N - 39.2 N = 10.8 N. Acceleration = F/m = 10.8 N / 10 kg = 1.08 m/s², approximately 1 m/s².
Q. A car is moving on a circular track of radius 100 m. If the maximum speed at which it can move without skidding is 20 m/s, what is the coefficient of friction between the tires and the road?
A.0.1
B.0.2
C.0.3
D.0.4
Solution
The centripetal force required is provided by friction: F = mv^2/r. The frictional force is μmg. Setting them equal gives μ = v^2/(rg). Here, μ = (20^2)/(100*9.8) ≈ 0.4.
Q. A car is negotiating a curve of radius 100 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent the car from skidding?
A.0.15
B.0.25
C.0.30
D.0.35
Solution
Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15 m/s)² / (100 m * 9.8 m/s²) ≈ 0.23.
