Answer to Question #166561 in Physics for David Samson

Question

a block of 1.5 kg is pushed up a rough 25 degrees inclined plane by a constant force of 15 newton acting parallel to the inclined plane. if the coefficient of the kinetic friction is 0.3 and the block is displaced by a distance 10cm along the inclined plane, calculate the work done against the force of gravity and against the force of friction

Answer 1

a) Work done against the force of gravity can be found as follows:

$$W_g=mgh.$$

We can find $h$ from the geometry:

$$sin\theta=\dfrac{h}{d},$$ $$h=dsin\theta.$$

Finally, we get:

$$W_g=mgdsin\theta=1.5\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot0.1\ m\cdot sin25^{\circ}=0.62\ J.$$

b) We can find the work done against the force of friction as follows:

$$W_{fr}=F_{fr}d=mgd \mu cos\theta,$$ $$W_{fr}=1.5\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot0.1\ m\cdot0.3\cdot cos180^{\circ}=-0.441\ J.$$