Question

A 1000 kg car pushes a 2000 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500N. Rolling friction can be neglected. A) What is the magnitude of the force of the car on the truck?B? What is the magnitude of the force of the truck on the car?

Answer 1

We will have the following:

We remember that:

`F=m\cdot a`

And we are given that:

`\begin{cases}m_c=1000kg \\ \\ m_t=2000kg \\ \\ F_T=4500N \\ \\ m_T=m_c+m_t\end{cases}`

Now, we will first determine the acceleration [The acceleration is shared between the car and the truck]:

`F=m_T\cdot a\Rightarrow a=(F)/(m_T)\Rightarrow a=(F)/(m_c+m_t)`
`\Rightarrow a=(4500N)/(1000kg+2000kg)\Rightarrow a=1.5m/s^2`

Now:

a) The magnitude of the force of the car on the truck is:

`F_t=m_t\cdot a\Rightarrow F_t=(2000kg)(1.5m/s^2)`
`\Rightarrow F_t=3000N`

So, the magnitude of the force of the car on the truck is 3000 N.

b) The magnitude of the force of the truck on the car is:

`F_c=m_c\cdot a\Rightarrow F_c=(1000kg)(1.5m/s^2)`
`\Rightarrow F_c=1500N`

So, the magnitude of the force of the truck on the car is 1500 N.