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