Question

A rough slope of length 5m is inclined at an angel of 30º to the horizontal. A body of
mass 2kg is released from the top of the slope and travels down the slope against a
constant resistance. The body reaches the bottom of the slope with speed 2 m/s.
Find the work done against the resistance.

Answer 1

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Answer on Question #50136, Physics, Mechanics | Kinematics | Dynamics

Task:

A rough slope of length 5m is inclined at an angle of $30{}^{\circ}$ to the horizontal. A body of mass 2kg is released from the top of the slope and travels down the slope against a constant resistance. The body reaches the bottom of the slope with speed 2 m/s. Find the work done against the resistance.

Answer:

Component of weight acting down slope = mgsin × 30 = 10N.

Resultant force down slope = ma = Component of weight acting down slope - friction

$$V^2 = U^2 + 2as, \quad U = 0, \quad so \quad a = V^2 / 2s.$$

Work done against the resistance = (Component of weight acting down slope - Resultant force down slope) * s = 50J - (m V²/2s) * s = 50J - 4J = 46J.