Question

A mass OF 5Kg is pulled along a rough inclined plane by a force in a direction making an angle 45° with the horizontal .The body moves with a uniform velocity through a distance 100m.Find the work done by the force .(co-efficient of friction =0.3)

Answer 1

$\Large\boxed{\quad\sf{4.6\ kJ}\quad}$

Given,

• $\sf{m=5\ kg}$

• $\sf{\theta=45^o}$

• $\sf{s=100\ m}$

• $\sf{\mu=0.3}$

Since the body moves with uniform velocity, the net acceleration as well as the net force acting on the body becomes zero by pulling it upward with the force F (say).

• Due to force of gravity, a force is acting on the body downwards along the plane surface and is equal to $\sf{mg\sin\theta.}$

• As the body is pulled upwards, frictional force is acting on it downwards along the plane surface and is equal to $\sf{\mu mg\cos\theta}$ where $\sf{mg\cos\theta}$ is the normal reaction on the block.

Therefore, since the net force is zero,

$\longrightarrow\sf{F=mg\sin\theta+\mu mg\cos\theta}$

$\longrightarrow\sf{F=mg[\sin\theta+\mu\cos\theta]}$

$\longrightarrow\sf{F=50[\sin45^o+0.3\cos45^o]}$

$\longrightarrow\sf{F=50\left[\dfrac{1}{\sqrt2}+\dfrac{0.3}{\sqrt2}\right]}$

$\longrightarrow\sf{F=\dfrac{50\times1.3}{\sqrt2}}$

$\longrightarrow\sf{F=32.5\sqrt2\ N}$

Then the work done by the force is,

$\longrightarrow\sf{W=Fs}$

[Angle between force and displacement is 0°, and hence cos 0° = 1.]

$\longrightarrow\sf{W=32.5\sqrt2\times100}$

$\longrightarrow\sf{\underline{\underline{W=3250\sqrt2\ J}}}$

$\longrightarrow\sf{\underline{\underline{W=4.6\ kJ}}}$