# Pushing a grocery cart with a force with a magnitude of 95 N, applied at an angle of 35° down from the cart horizontal, makes the cart travel at a constant speed of 1.2 m/s. What is the frictional force acting on the cart?

Jul 12, 2018

$\text{friction} \cong 78 N$

#### Explanation:

Since the speed is constant, there is no acceleration. Since there is no acceleration, the net force acting on the cart is zero. In the horizontal direction, there are only 2 forces action on the cart: the horizontal component of the 95 N (in the forward direction)and the friction (in the rearward direction). They combine to form the net force as follows.

${F}_{\text{net" = 95 N*cos35^o - "friction}}$

Since I said the net force is zero because of constant velocity

$95 N \cdot \cos {35}^{o} - \text{friction} = 0$

$\text{friction} = 95 N \cdot \cos {35}^{o} = 77.82 N \cong 78 N$

I hope this helps,
Steve