Question

A shopper pushes a grocery cart 41.9 m on level ground, against a 44.5 N frictional force. The cart has a mass of 16.3 kg. He pushes in a direction 17.5º below the horizontal. If the initial velocity was 1.9 m/s, and final speed was 12.6 m/s. What is the work done by the pushing force in unit of Joule?

Answer 1

Answer:

Fp = 26.59[N]

Explanation:

This problem can be solved using the principle of work and energy conservation, i.e. the final kinetic energy of a body will be equal to the sum of the forces that do work on the body plus the initial kinetic energy.

We need to identify the initial data:

d = distance = 41.9[m]

Ff = friction force = 44.5 [N]

m = mass = 16.3 [kg]

v1 = 1.9 [m/s]

v2 = 12.6 [m/s]

The kinetic energy at the beginning can be calculated as follows:

$E_{k1}= \frac{1}{2}*m*v_{1}^2 \\E_{k1}= \frac{1}{2}*16.3*(1.9)_{1}^2\\E_{k1}= 29.42[J]$

And the final kinetic energy.

$E_{k2}= \frac{1}{2}*m*v_{2}^2 \\E_{k2}= \frac{1}{2}*16.3*(12.6)^2\\E_{k2}= 1294[J]$

The work is performed by two forces, the friction force and the pushing force, it is important to clarify that these forces are opposite in direction.

The weight of the cart also performs a work in the direction of movement since the plane is tilted down, this component of the weight of the cart must be parallel to the surface of the inclined plane.

$W_{1-2}=-(44.5*41.9)+(16.3*9.81*sin(17.5)*41.9)+(F_{p}*41.9) \\therefore:\\E_{k1}+W_{1-2}=E_{k2}\\29.42+150.16+(F_{p}*41.9)=1294\\F_{p}=1114.42/41.9\\F_{p}=26.59[N]$