Question

A student pulls his younger sister in a sled (combined mass 31 kg) with a rope that is angled at = 13° with a force of 290 N. The coefficient of friction between the sled and the ground is 0.28.What is the acceleration of the sled?

Answer 1

6.37 m/s²

Step-by-step explanation:

We can represent the situation using the following diagram

First, we need to find the normal force. Since the sled is not moving upwards, the net force in the vertical direction, we get that the normal force is equal to the weight so

Now, there is a net force in the horizontal direction and by the second law of newton, it is equal to

`\begin{gathered} F_{\text{net}}=F\cos \theta-F_f=ma \\ F\cos \theta-\mu F_n=ma \end{gathered}`

Where Fcos is the horizontal component of F, μ is the coefficient of friction, Fn is the normal force and a is the acceleration. So, solving for a, we get

`a=(F\cos \theta-\mu F_n)/(m)`

Now, we can replace F = 290 N, = 13°, μ = 0.28, Fn = 303.8N and m= 31 kg to get

Therefore, the acceleration of the sled is 6.37 m/s²