Question

A 1280-kg car pulls a 350-kg trailer. The car exerts a horizontal force of 3.6 × 10^3 N against the ground in order to accelerate. What force does the car exert on the trailer? Assume an effective friction coefficient of 0.15 for the trailer.

Answer 1

1176 N

Step-by-step explanation:

We are given that

Mass of car=
`m_1=`1280 kg

Mass of trailer=
`m_2`=350 kg

Car exerts a horizontal force against the ground =
`3.6* 10^(3)`N

Coefficient of friction=
`\mu=0.15`

We have to find the force exerted by car on the trailer.

`F=(m_1+m_2)a+F_f`

`F=(m_1+m_2)a+\mu m_2g`

`g=9.8 m/s^2`

Substitute the values then we get

`3.6* 10^(3)=(1280+350)a+0.15* 350* 9.8`

`3.6* 10^(3)=1630a+514.5`

`1630a=3600-514.5=3085.5`

`a=(3085.5)/(1630)=1.89 m/s^2`

Force exert on the trailer

`F=m_2a+\mu m_2g`

`F=350(1.89)+0.15* 350* 9.8`

`F=1176 N`

Hence, the car exerts force on the trailer=1176 N