Please help me with question number 6 part (i)?

A car of mass 1250 kg moves from the bottom to the top of a straight hill of length 500 m. The top of the hill is 30 m above the level of the bottom. The power of the car’s engine is constant and equal to 30 000 W. The car’s acceleration is #4 ms^2# at the bottom of the hill and is #0.2 ms^2# at the top.

The resistance to the car’s motion is 1000 N. Find

(i)

the car’s gain in kinetic energy,

(ii)

the work done by the car’s engine.

1 Answer
A08
May 15, 2018

Driving force #F = "Power"/"velocity"=30000/v \ N#
At the bottom of hill.
Forces acting on the car along the hill.

#Fuarr+sin theta" component of weight"darr+"resistance"darr# .....(1)

From Newton's Second Law of motion

#F_"net"=1250xx4=5000# .....(2)

Let #v_b# be the velocity at the bottom of hill. Equating (1) and (2) and taking #g=10\ ms^-2# we get

#30000/v_b-1250g(30/500)-1000=5000#
#=>30000/v_b-750-1000=5000#
#=>v_b=4.bar4\ ms^-1#

Similarly, if #v_t# be velocity at the top of hill we get, (#F_"net"# in equation (2) changes due to change in acceleration.)

#30000/v_t-750-1000=1250xx0.2=250#
#=>v_t=15\ ms^-1#

Gain in KE

#DeltaKE=1/2mv_t^2-1/2mv_b^2#
#=>DeltaKE=1/2xx1250((15)^2-(4.bar4)^2)#
#=>DeltaKE=128279\ J#

(ii) Using Law of conservation of Energy

#"Work done by the car’s engine " W_T = "Gain in KE" + "Gain in PE" + "Work done against resistance"#
#=>W_T = 128279 + mgh + vecF_rcdotvecd#
#=>W_T = 128279 + 1250xx10xx30 + 1000xx500#
#=>W_T = 128279 + 375000 + 500000#
#=>W_T=1003279\ J#

Related questions
Impact of this question
3215 views around the world
You can reuse this answer
Creative Commons License