A block of mass 1 kg is pushed up a rough surface that is inclined to the horizontal at an angle of $30^\circ$. The push is provided by a force of 10 N directed parallel to the inclined surface, up the slope. The coefficient of friction between the block and the incline is 0.1. The block is pushed up through a distance of 10 m measured along the incline. Taking $g = 9.8\ \text{m s}^{-2}$, calculate (a) the work done against gravity, (b) the work done against the force of friction, (c) the increase in potential energy, (d) the increase in kinetic energy, and (e) the work done by the applied force.

> [!TLDR] > The block is pushed 10 m up a $30^\circ$ rough incline by a 10 N force. Work against gravity $=mgh=49$ J, work against friction $=\mu mg\cos30^\circ\times d\approx 8.5$ J, PE gain $=49$ J, and by the work–energy theorem the KE gain $=$ applied work (100 J) $-49$ J $-8.5$ J $\approx 42.5$ J. The applied force does 100 J. ### Given - $m = 1\ \text{kg}$, incline angle $\theta = 30^\circ$, applied force $F = 10\ \text{N}$ (up the incline), $\mu = 0.1$, displacement along incline $d = 10\ \text{m}$, $g = 9.8\ \text{m s}^{-2}$. ### (a) Work done against gravity Vertical height gained: $h = d\sin\theta = 10 \times \sin 30^\circ = 5\ \text{m}$. $$W_{grav} = mgh = 1 \times 9.8 \times 5 = 49\ \text{J}$$ ### (b) Work done against friction Normal reaction $N = mg\cos\theta$, so the friction force is $f = \mu N = \mu mg\cos\theta$. $$f = 0.1 \times 1 \times 9.8 \times \cos 30^\circ = 0.1 \times 9.8 \times 0.866 = 0.849\ \text{N}$$ $$W_{fric} = f\,d = 0.849 \times 10 \approx 8.5\ \text{J}$$ ### (c) Increase in potential energy $$\Delta PE = mgh = 49\ \text{J}$$ This equals the work done against gravity. …

A block of mass 1 kg is pushed up a rough surface that is inclined to the horizontal at an angle of 30^°. The

NCERT Exemplar · Q44

30^\circ

(a) the work done against gravity,

(b) the work done against the force of friction,

(d) the increase in kinetic energy, and

(e) the work done by the applied force.

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Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

Imagine pushing a heavy box across the floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — the push (force) over a distance (displacement) changing the box's speed — is exactly what the Work-Energy Theorem captures.

The Intuition First

Think of work as the "currency" that buys motion. When you do work on an object, you transfer energy to it. That energy shows up as kinetic energy — the energy of motion. The more work you do, the more the object's kinetic energy changes.

If you push a stationary ball, it starts moving. If you push a moving ball in the same direction, it speeds up. If you push against its motion, it slows down. In every case, the work done equals the change in the ball's kinetic energy.

Note

Work is done by a force on an object. The object's kinetic energy changes by exactly that amount (assuming no other forces do work).

The Precise Statement

W_{\text{net}} = \Delta K = K_f - K_i

Where:

$W_{\text{net}}$ is the net work done on the object (the total work from all forces combined)
$K_f$ is the final kinetic energy
$K_i$ is the initial kinetic energy

And kinetic energy is defined as:

K = \frac{1}{2}mv^2

So the theorem can also be written as:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Why "Net" Work Matters

This is the most common point of confusion. The theorem uses net work — the work done by the net force (the vector sum of all forces). If you push a box and friction opposes it, the net work is the work you do minus the work friction does. Only that net amount changes the kinetic energy.

Watch out

If you push a box at constant speed, your work is positive, but friction does equal negative work. The net work is zero, so kinetic energy doesn't change — the box keeps moving at the same speed. Your work didn't "disappear"; it was dissipated as heat by friction.

A Simple Derivation (for constant force)

Consider a constant net force $F_{\text{net}}$ acting on an object of mass $m$ over a displacement $s$ . From Newton's second law:

F_{\text{net}} = ma

From kinematics (constant acceleration):

v_f^2 = v_i^2 + 2as

Multiply both sides by $\frac{1}{2}m$ :

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + mas

But $mas = F_{\text{net}} s = W_{\text{net}}$ , so:

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + W_{\text{net}}

Rearranging:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta K

Tip

The theorem holds even for variable forces and curved paths — the derivation uses calculus then, but the result is the same.

What It Tells You (and What It Doesn't) …

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The block is pushed 10 m up a $30^\circ$ rough incline by a 10 N force. Work against gravity $=mgh=49$ J, work against friction $=\mu mg\cos30^\circ\times d\approx 8.5$ J, PE gain $=49$ J, and by the work–energy theorem the KE gain $=$ applied work (100 J) $-49$ J $-8.5$ J $\approx 42.5$ J. The applied force does 100 J.

Given

$m = 1\ \text{kg}$ , incline angle $\theta = 30^\circ$ , applied force $F = 10\ \text{N}$ (up the incline), $\mu = 0.1$ , displacement along incline $d = 10\ \text{m}$ , $g = 9.8\ \text{m s}^{-2}$ .

(a) Work done against gravity

Vertical height gained: $h = d\sin\theta = 10 \times \sin 30^\circ = 5\ \text{m}$ .

W_{grav} = mgh = 1 \times 9.8 \times 5 = 49\ \text{J}

(b) Work done against friction

Normal reaction $N = mg\cos\theta$ , so the friction force is $f = \mu N = \mu mg\cos\theta$ .

f = 0.1 \times 1 \times 9.8 \times \cos 30^\circ = 0.1 \times 9.8 \times 0.866 = 0.849\ \text{N}

W_{fric} = f\,d = 0.849 \times 10 \approx 8.5\ \text{J}

(c) Increase in potential energy

\Delta PE = mgh = 49\ \text{J}

This equals the work done against gravity. …

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Given

(a) Work done against gravity

(b) Work done against friction

(c) Increase in potential energy

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