Question

A grocery cart of mass 16 kg is being pushed at a constant speed up a 12-degree ramp by a force of [{MathJax fullWidth='false' f_p }] which acts at an angle of 17 degrees below the horizontal. Find the work done by each of the forces ([{MathJax fullWidth='false' m_g, f_n, f_p }]) on the cart if the ramp is 7.5 m long.

Answer 1

Answer:

If the cart is being pushed at a constant speed, then the acceleration in the direction of motion is zero. Hence, the force in the direction of the motion is zero, according to Newton's Second Law.

$\Sigma F_x = ma_x$

For simplicity, I will denote the direction along the inclined ramp as x-direction.

In the question the value of the force is not clearly given, so I will denote it as $F_P$

$\Sigma F_{net_x} = ma_x\\\Sigma F = F_{P}\cos(29^\circ) - mg\sin(12^\circ) = ma_x = 0\\F_{P}\cos(29^\circ) = mgsin(12^\circ)\\F_{P}\times 0.8746 = 16\times 9.8\times 0.2079\\F_{P} = 37.2740$

Here the angle between the applied force and the x-direction is 12° + 17° = 29°

The x-component of the weight of the cart is equal to sine component of the weight.

Since the cart is rolling on tires the kinetic friction does no work.

Work done by the applied force:

$W_{F_P} = F_P_x \cos(29^\circ)\times 7.5 = 244.5 ~J$

Work done by the weight of the cart:

$W_{mg} = -mg\sin(12^\circ)\times 7.5 = -16\times 9.8 \times 0.2079 \times 7.5 = - 244.5~J$

Since the x-component of the weight is in the -x-direction, its work is negative.

Conveniently, the total work done on the particle is zero, since its velocity is constant.