A shopper pushes a grocery cart 20.0 m at constant speed

Step 1 of 5

The man pushes the cart 20 m in a direction 250 below the horizontal at a constant speed against a 35 N friction force.

a) We need to find out the work done on the cart by friction.

 The work done by the friction is in the opposite direction of motion.

So the work done by friction is given by

\(W_{f}=F d \operatorname{Cos} \theta\)

The angle \(\theta=180^{\circ}\)

So \(W_{f}=F d \operatorname{Cos}\left(180^{\circ}\right)\)

F is the frictional force \(F=35 N\)

d is the distance, \(d=20 m\)

\(\begin{array}{l}
W_{f}=35 \mathrm{~N} \times 20 \mathrm{~m} \\
W_{f}=700 \mathrm{~J}
\end{array}\)