Question

A 3.00 kg cart on a track is pulled by a string so that it accelerates at 2.00 m/s/s. The force of tension in the string is 10.0 N. What is the force of friction on the cart?

Answer 1

Answer:

If the track is horizontal and the string is pulled horizontally, the friction on the cart would be $4.0\; \rm N$.

Explanation:

Let $m$ denote the mass of this cart, and let $a$ denote the acceleration of this cart.

$m = 3.00\; \rm kg$.

$a = 2.00\; \rm m \cdot s^{-2}$.

Apply Newton's Second Law to find the net force on this cart.

$\begin{aligned}\text{Net Force} &= m \cdot a\\ &= 3.00\; \rm kg \times 2.00\; \rm m\cdot s^{-2}\\ &= 6.00\; \rm N\end{aligned}$.

The following forces act upon this cart:

• (downward) gravitational attraction from the earth,
• (upward) normal force from the track,
• (forward) tension from the string, and
• (backward) friction from the track.

Assume that the track is horizontal, and that the string was pulled horizontally. The normal force from the track would exactly balance the downward gravitational attraction from the earth. Hence, the $6.00\; \rm N$ net force on this cart would be equal (in size) to the size of the tension from the string ($10.0\; \rm N$) minus the size of the friction from the track.

In other words:

$\begin{aligned}&\text{Size of Net Force}\\ &= \text{Size of Tension} - \text{size of friction}\end{aligned}$.

$\begin{aligned}& 6.00\;\rm N = 10.0\; \rm N - (\text{size of friction})\end{aligned}$.

$\text{size of friction} = 10.0\; \rm N - 6.00\; \rm N = 4.0\; \rm N$.