Question

Mechanic help?

Answer 1

This is what I get

Explanation:

(i) Power `P` is given by

`P=vecFcdotvecv`
where `vecF` is the force applied and `vecv` is the velocity of the object.

The van's engine is required to maintain a velocity of `25\ ms^-1` along the horizontal direction and over come total resistance force `=300("van")+100("trailer")=400\ N`.
As force applied to overcome the resistance and velocity are along the same direction, angle between the two `=0=>cos0^@=1`

`:.P=400xx25=10000\ W`

(ii) Tension in the cable is equal to the resistance to the motion of the trailer.

`T=100\ N`

(iii) As the van pulls the trailer uphill, power of engine is used to overcome sum of resistances to the motion of the trailer and van combine and sin theta components of their weights which pulls the van trailer combination back. Now force applied by engine can be found from

`25000=|vecF_"up incline"|xx20xxcos0^@`
`=>F_"up incline"=25000/20`
`=>F_"up incline"=1250\ N`

Let `a_"up"` be net acceleration at that instant. The force equation can be written as

`(3000+500)a_"up"=1250-(3000+500)xx9.81xxsin4^@-300-100`
`=>a_"up"=850/(3500)-9.81xx0.06976`
`=>a_"up"=-0.44\ ms^-2`
`:.T_"up"="mass of trailer"xxa_"up"`
`=>T_"up"=221.4\ N`