A block of mass 1kg is pushed up a surface inclined to horizontal at an angle of $30^{\circ}$ by a force of 10N parallel to the inclined surface $[f i g u r e]$ . The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate
(a) work done against gravity
(b) work done against force of friction
(c) increase in potential energy
(d) increase in kinetic energy
(e) work done by applied force.

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In figure, f is the force of friction acting down the incline.
Here, $m = 1 k g, θ = 30^{\circ}$
$F = 10 N, μ = 0.1, s = 10 m$ . (a) Work done against gravity
$W_{g} = m g sin θ \times s = 1 \times 10 {sin 30}^{\circ} \times 10 = \frac{100}{2} = 50 J$
(b) Work done against friction
$W_{f} = f \times s = μ R \times s = μ m g cos θ \times s$
$= 0.1 \times 1 \times 10 {cos 30}^{\circ} \times 10 = 10 \times 0.866 = 8.66 J$
(c) Increase in P.E. $= m g h = m g (s sin θ) = 1 \times 10 \times 10 {sin 30}^{\circ} = 50 J$
(d) Increase in K.E. $= Δ K = ?$
Acceleration of block up the incline
$a = \frac{F - (m g sin θ + μ m g cos θ)}{m} = \frac{10 - (1 \times 10 \times \frac{1}{2} + 0.1 \times 10 \times 0.866)}{1} = 4.13 m / s^{2}$
From $υ^{2} - u^{2} = 2 a s$
$Δ K = \frac{1}{2} m (υ^{2} - u^{2}) = \frac{1}{2} m \times 2 a s = m a s = 1 \times 4.13 \times 10 = 41.3 J$

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Updated on:21/07/2023

Class 11 PHYSICS WORK, ENERGY AND POWER

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