Questions & explanations
1. A 2 kg block is placed on a 8 kg wedge of angle 30°. The wedge is pushed horizontally with 50 N on smooth ground. Coefficient of friction between block and wedge is 0.3. Find the acceleration of the wedge if the block does not slip. (g = 10 m/s²)
- 4.0 m/s²
- 5.0 m/s²
- 6.0 m/s²
- 3.0 m/s²
Answer: 5.0 m/s²
If block does not slip, both move together. Net force on system = 50 N, total mass = 10 kg, so acceleration a = 50/10 = 5.0 m/s². Check friction: pseudo-force on block = 2×5=10 N horizontally. Normal N = mg cos30° + ma sin30° = 20×0.866 + 10×0.5 = 17.32+5=22.32 N. Required friction along incline = ma cos30° - mg sin30° = 10×0.866 - 20×0.5 = 8.66-10 = -1.34 N (upward). Maximum static friction = μN = 0.3×22.32=6.7 N > 1.34 N, so no slip.
2. A block of mass 2 kg is placed on a block of mass 4 kg. The lower block is on a rough floor (μ_k=0.2). The upper block is pulled by a force F=30 N at 30° above horizontal. The coefficient of static friction between blocks is 0.5. Find the acceleration of the system if the blocks move together. (g=10 m/s²)
- 2.5 m/s²
- 3.0 m/s²
- 1.5 m/s²
- 4.0 m/s²
Answer: 2.5 m/s²
For the system to move together, treat both blocks as one object of mass 4 kg. Horizontal force = 20 N. Normal force from floor = 40 N, so kinetic friction = 0.2 × 40 = 8 N. Net force = 20 - 8 = 12 N. Acceleration = 12 / 4 = 3.0 m/s². Check if they actually move together: required friction on lower block = 2 × 3 = 6 N, which is less than maximum static friction between blocks (0.5 × 20 = 10 N), so they do move together.
3. A block of mass m slides down a smooth wedge of mass M on a smooth floor. In the ground frame, the wedge accelerates left. What is the horizontal acceleration of the block relative to ground?
- g sinθ cosθ
- a_wedge - g sinθ cosθ
- a_wedge + g sinθ cosθ
- g sinθ cosθ - a_wedge
Answer: g sinθ cosθ - a_wedge
Using Newton's second law in ground frame: block's horizontal acceleration = wedge's acceleration (a_wedge) plus relative acceleration along incline's horizontal component. The relative acceleration down incline is g sinθ, so horizontal component = g sinθ cosθ. Since wedge moves left (negative direction) and block moves right relative to wedge, net horizontal acceleration = g sinθ cosθ - a_wedge.
4. A block of mass 5 kg is on a rough incline of angle 30° with μ_s = 0.4. A horizontal force F is applied to keep the block from sliding down. What is the minimum value of F? (g = 10 m/s²)
- 8.7 N
- 12.5 N
- 10.2 N
- 15.3 N
Answer: 8.7 N
For impending motion down the incline, friction acts up the incline. Equilibrium: N = mg cosθ + F sinθ, mg sinθ = F cosθ + μ_s N. Solving gives F = mg (sinθ - μ_s cosθ)/(cosθ + μ_s sinθ). With m=5 kg, θ=30°, μ_s=0.4, g=10, we get F = 50×(0.5 - 0.4×0.866)/(0.866 + 0.4×0.5) = 50×0.1536/1.066 = 7.2 N. Closest option is 8.7 N.
5. A car takes a turn of radius 60 m on a banked road of angle 30° with coefficient of friction 0.2. What is the maximum safe speed? (g = 10 m/s²)
- √[600(0.577-0.2)/(1-0.2*0.577)] ≈ 17.3 m/s
- √[600(0.577-0.2)/(1+0.2*0.577)] ≈ 14.9 m/s
- √[600(0.577+0.2)/(1+0.2*0.577)] ≈ 19.8 m/s
- √[600(0.577+0.2)/(1-0.2*0.577)] ≈ 22.4 m/s
Answer: √[600(0.577+0.2)/(1-0.2*0.577)] ≈ 22.4 m/s
For maximum safe speed on a banked rough road, friction acts down the slope. The formula is v_max = √[rg(tanθ+μ)/(1-μtanθ)]. With r=60, g=10, tan30°=0.577, μ=0.2, we get v_max = √[600(0.577+0.2)/(1-0.2*0.577)] = √[600*0.777/0.8846] ≈ √[527] ≈ 22.4 m/s. Option d matches this formula and value.
6. Three blocks of masses 2 kg, 3 kg, and 5 kg are pushed by a 50 N force on a rough floor with μ_k = 0.2. Find the contact force between the 2 kg and 3 kg blocks. (g = 10 m/s²)
- 20 N
- 30 N
- 50 N
- 40 N
Answer: 40 N
Total mass = 10 kg, total friction = 0.2 × 10 × 10 = 20 N. Net force = 50 - 20 = 30 N, common acceleration a = 30/10 = 3 m/s². Isolate blocks 2 and 3: mass = 8 kg, friction on them = 0.2 × 8 × 10 = 16 N. Net force needed = 8 × 3 = 24 N. Contact force = net force + friction = 24 + 16 = 40 N.
7. For an isolated system of interacting particles, what does Newton's third law imply about the total momentum?
- Total momentum decreases with time
- Total momentum increases with time
- Total momentum is conserved
- Total momentum oscillates
Answer: Total momentum is conserved
Newton's third law states that internal forces between particles are equal and opposite. Summing over all particles, the net internal force is zero. For an isolated system, no external force acts, so the rate of change of total momentum is zero, meaning total momentum remains constant.
8. Two blocks of masses 2 kg and 3 kg are stacked on a smooth floor. A horizontal force of 20 N is applied to the lower block. The coefficient of static friction between the blocks is 0.5. What is the acceleration of the upper block?
- 6 m/s²
- 5 m/s²
- 2 m/s²
- 4 m/s²
Answer: 4 m/s²
Assume both blocks move together. Common acceleration a = F/(m1+m2) = 20/5 = 4 m/s². Required friction on upper block f_req = m1 a = 2×4 = 8 N. Maximum static friction f_max = μ m1 g = 0.5×2×10 = 10 N. Since 8 N ≤ 10 N, no slipping; upper block accelerates at 4 m/s².
9. A block of mass m is placed on a wedge of mass M on a smooth floor. The coefficient of friction between block and wedge is μ. What is the acceleration of the wedge just before the block slips down?
- g (μ cosθ - sinθ) / (cosθ + μ sinθ)
- g (sinθ - μ cosθ) / (cosθ + μ sinθ)
- g (μ cosθ + sinθ) / (cosθ - μ sinθ)
- g (μ sinθ - cosθ) / (sinθ + μ cosθ)
Answer: g (sinθ - μ cosθ) / (cosθ + μ sinθ)
Using Newton's second law in the wedge's non-inertial frame, pseudo-force ma acts on block. Resolving forces along incline: mg sinθ + ma cosθ = μN (impending slip). Perpendicular: N = mg cosθ - ma sinθ. Substituting N gives a = g (sinθ - μ cosθ) / (cosθ + μ sinθ).
10. A car takes a turn of radius 50 m on a banked road with angle θ = 15° and μ_s = 0.4. What is the maximum safe speed? (tan15° ≈ 0.27, g = 10 m/s²)
- 19.4 m/s
- 14.1 m/s
- 22.4 m/s
- 17.3 m/s
Answer: 19.4 m/s
Maximum safe speed on banked rough road: v_max = √(rg (tanθ + μ_s)/(1 - μ_s tanθ)). Plug values: r=50, g=10, tanθ=0.27, μ_s=0.4. Numerator = 0.27+0.4=0.67, denominator = 1 - 0.4×0.27 = 0.892. v_max = √(50×10×0.67/0.892) = √(335/0.892) ≈ √375.6 ≈ 19.4 m/s.
11. A road is banked at an angle θ for a speed v. Which graph correctly shows the variation of tanθ with v² for a fixed radius r?
- A curve that flattens at high v²
- A straight line through origin with slope rg
- A straight line with positive intercept on tanθ axis
- A straight line through origin with slope 1/(rg)
Answer: A straight line through origin with slope 1/(rg)
For a banked road without friction, the centripetal force is provided by the horizontal component of the normal reaction: N sinθ = mv²/r and N cosθ = mg. Dividing gives tanθ = v²/(rg). Thus tanθ vs v² is a straight line through origin with slope 1/(rg).
12. A block of mass 5 kg on a rough horizontal surface has μ_s = 0.6. What is the minimum force required to move the block? (g = 10 m/s²)
- 30 N
- 18 N
- 25.7 N
- 50 N
Answer: 25.7 N
Minimum force to move a block on a rough horizontal surface is achieved by applying force at an angle. Using the formula F_min = μ_s mg / √(1+μ_s²) = 0.6×5×10 / √(1+0.36) = 30/√1.36 ≈ 25.7 N. This is less than the horizontal pull of 30 N.