Answer to Question #116397 in Quantum Mechanics for Foibe Kambala

Question

6. What is the minimum work needed to push a 950 kg car 810 m up along a 9.0° incline?
a) Ignore friction
b) Assume the effective coefficient of friction retarding the car is 0.25

Answer 1

We have given,

mass of the car $m=950Kg$ ,inclined distance $d=810m$ and the angle of inclination $\theta=9.0^{\circ}$ .

a).Let's draw the FBD of the system,

clearly from the above free body diagram (FBD), to lift up the car along the inclined plane,

minimum applied force must be,

$$F_{applied}=mg\sin(\theta)$$

Thus, work done by the $F_{applied}$ will be,

$$W_{applied}=F_{applied}\cdot d$$

Hence,

$$W_{applied}=mgd\sin(\theta)\\ \implies W_{applied}=950\times 9.8\times810\times \sin(9.0^{\circ})\\ \implies W_{applied}=11,79,687.94J$$

b).In this case frictional force is acting and coefficient of friction $\mu=0.25$ given.

Again draw the FBD of the system,

From the above FBD, we get that the minimum applied force need to be,

$$F_{applied}=mg\sin(\theta)+f$$

Since,

$$f=\mu N$$

where, $N$ is the normal force acting between incline plane and the given car,thus

$$N=mg\cos(\theta)\implies f=\mu mg\cos(\theta)$$

Therefore,

$$F_{applied}=mg(\sin(\theta)+\mu \cos(\theta))\\ \implies W_{applied}=F_{applied}\cdot d\\ \implies W_{applied}=mgd(\sin(\theta)+\mu \cos(\theta))\\ \implies W_{applied}=950\times9.8\times810\times(\sin(9.0^{\circ})+0.25\cos(9.0^{\circ}))\\ \implies W_{applied}=30,41,752.08J$$

Therefore we are done.