Question

A block of mass 20 kg is slowly slid up on a smooth incline of inclination 53° by a person. Calculate the work done by the person in moving the block through a distance of 4 m, if the driving force is (a) parallel to the incline and (b) in the horizontal direction. [g = 10 m/s²]

Answer 1

640 J in both cases

Answer 2

The problem asks for the work done by the person to slowly slide a block of mass $m=20$ kg up a smooth incline of inclination $\theta=53^\circ$ through a distance $d=4$ m. The process is slow, meaning the acceleration is negligible, so the net force along the incline is zero. The acceleration due to gravity is $g=10$ m/s². We use $\sin 53^\circ \approx 0.8$ and $\cos 53^\circ \approx 0.6$.

The displacement of the block is along the incline, upwards, with magnitude $d=4$ m.

(a) Driving force is parallel to the incline.

Let the driving force be $\vec{F}_{p,a}$. Since the block is moved up slowly, the net force along the incline is zero. The forces acting along the incline are the driving force $\vec{F}_{p,a}$ upwards and the component of gravity along the incline, $mg \sin \theta$, downwards.

$F_{p,a} - mg \sin \theta = 0$

$F_{p,a} = mg \sin \theta = (20 \text{ kg})(10 \text{ m/s}^2)\sin 53^\circ = 200 \times 0.8 = 160$ N.

The driving force is parallel to the incline and in the direction of displacement. The angle between the force and displacement is $0^\circ$.

The work done by the person is $W_a = \vec{F}_{p,a} \cdot \vec{d} = F_{p,a} d \cos 0^\circ = F_{p,a} d$.

$W_a = (160 \text{ N})(4 \text{ m}) = 640$ J.

(b) Driving force is in the horizontal direction.

Let the driving force be $\vec{F}_{p,b}$. This force is horizontal. To move the block up the incline, the horizontal force must have a component along the incline. Let the magnitude of the horizontal force be $F_{p,b}$.

We resolve the forces along the incline and perpendicular to the incline.

The displacement is along the incline, upwards. The incline makes an angle $\theta$ with the horizontal. The horizontal force $\vec{F}_{p,b}$ makes an angle $\theta$ with the incline. The component of $\vec{F}_{p,b}$ along the incline is $F_{p,b} \cos \theta$. This component acts upwards along the incline.

The forces acting along the incline are the component of the driving force along the incline, $F_{p,b} \cos \theta$, upwards, and the component of gravity along the incline, $mg \sin \theta$, downwards.

Since the block is slid up slowly, the net force along the incline is zero:

$F_{p,b} \cos \theta - mg \sin \theta = 0$

$F_{p,b} \cos \theta = mg \sin \theta$

$F_{p,b} = mg \frac{\sin \theta}{\cos \theta} = mg \tan \theta = (20 \text{ kg})(10 \text{ m/s}^2)\tan 53^\circ = 200 \times \frac{0.8}{0.6} = 200 \times \frac{4}{3} = \frac{800}{3}$ N.

The work done by the person is $W_b = \vec{F}_{p,b} \cdot \vec{d}$. The force $\vec{F}_{p,b}$ is horizontal and the displacement $\vec{d}$ is along the incline. The angle between the horizontal direction and the incline is $\theta = 53^\circ$.

$W_b = F_{p,b} d \cos \theta$.

$W_b = \left(\frac{800}{3} \text{ N}\right) (4 \text{ m}) \cos 53^\circ = \frac{800}{3} \times 4 \times 0.6 = \frac{800}{3} \times 4 \times \frac{3}{5}$.

$W_b = \frac{800 \times 4 \times 3}{3 \times 5} = \frac{800 \times 4}{5} = 160 \times 4 = 640$ J.

Alternatively, we can calculate the work done by the person as the negative of the work done by gravity, since the net work is zero (as $\Delta K = 0$ and $W_N = 0$). The work done by gravity is $W_g = -mgh$, where $h$ is the vertical displacement.

The vertical displacement is $h = d \sin \theta = 4 \sin 53^\circ = 4 \times 0.8 = 3.2$ m.

$W_g = -(20 \text{ kg})(10 \text{ m/s}^2)(3.2 \text{ m}) = -640$ J.

Work done by the person is $W_p = -W_g = 640$ J. This is true regardless of the direction of the applied force, as long as the block is moved slowly up the incline by the person's force.

The work done by the person is 640 J in both cases.