A body of mass 20Kg is resting on a horizontal surface. A horizontal force F = 180N is applied to the body, and the body starts to slide. Calculate the acceleration of the body.

The coefficient of sliding friction = 0.5

The coefficient of static friction = 0.6

Use g=9.81 m/s2 in calculation.

Question

A body of mass 20Kg is resting on a horizontal surface. A horizontal force F = 180N is applied to the body, and the body starts to slide. Calculate the acceleration of the body.

The coefficient of sliding friction = 0.5

The coefficient of static friction = 0.6

Use g=9.81 m/s2 in calculation.

BlackTom AI · Accepted Answer

We start by identifying the forces on the block on a horizontal surface. The weight is W = mg = 20 × 9.81 = 196.2 N, and since the surface is horizontal, the normal force N equals W, so N = 196.2 N.

Static friction maximum is f_s,max = μ_s N = 0.6 × 196.2 ≈ 117.72 N. The applied horizontal force is F = 180 N, which exceeds f_s,max, so the block will begin to slide rather than stay at rest.

Once sliding begins, kinetic friction acts opposite to the motion with magnitude f_k = μ_k N = 0.5 × 196.2 = 98.1 N.

The net horizontal force causing acceleration once sliding starts is F_net = F − f_k = 180 − 98.1 = 81.9 N.

Using Newton's second law, a = F_net / m = 81.9 / 20 = 4.095 m/s^2.

Now considering the diagram showing a force at 30°, if the force were inclined at 30°, the horizontal component would be F_x = 180 cos 30° ≈ 155.9 N and the vertical component would reduce the normal force to N = mg − F_y = 196.2 − 180 sin 30° = 196.2 − 90 = 106.2 N. Then f_k = μ_k N = 0.5 × 106.2 ≈ 53.1 N, and the net horizontal force would be F_x − f_k ≈ 102.8 N, giving a ≈ 102.8/20 ≈ 5.14 m/s^2. However, the problem statement explicitly says the force is horizontal, so the horizontal-only calculation leading to 4.095 m/s^2 is the intended result.

Now evaluate the provided answer options:
- 3.114 m/s^2: This would correspond to a smaller net force or larger friction, not matching the computed net force with given numbers.
- 4.095 m/s^2: This exactly matches the a = (180 − 98.1)/20 computed for a horizontal applied force.
- 9 m/s^2: This would require a much larger net force or smaller mass than given; not consistent with the calculation.

Therefore, the option that aligns with the horizontal-force scenario and the calculations above is the second option, 4.095 m/s^2.

INU1127 (24/25) Friction - Quiz 1

A body of mass 20Kg is resting on a horizontal surface. A horizontal force F = 180N is applied to the body, and the body starts to slide. Calculate the acceleration of the body. The coefficient of sliding friction = 0.5 The coefficient of static friction = 0.6 Use g=9.81 m/s2 in calculation.

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