A 1500-kg car starts from rest to reach 60.0 miles/h in 7.50 s. How much work is
ID: 2257338 • Letter: A
Question
A 1500-kg car starts from rest to reach 60.0 miles/h in 7.50 s. How much work is done by the engine during the acceleration? What is the engine's power output? If the same car moving at 55.0 miles/h brakes to a complete stop, how much heat (in joules AND In kcals) is generated? 1 kcal = 4186 J. A 45.0-g bullet is shot straight upward with a muzzle speed of 350 m/s. Find the maximum height the bullet can reach by using two different approaches: one with kinematics equations, and the other by considering energy (assuming no air resistance). A toy gun can shoot 10.00-g plastic bullets at 5.00 m/s. If the spring in the gun can be compressed by 4.00 cm, what is its spring constant? What compression should be if the speed of the bullet is to be increased to 7.00 m/s?Explanation / Answer
Qsn 5) a) work done = change in K.E
= 0.5 x m x (Vf ^2 - Vi ^2) (Vi=0 Vf = 60 miles/hr => Vf = 60 x 1609.34 m/ 3600 s)
= 0.5 x 1500 x (26.82 ^2 - 0) => Vf = 26.82 m/s
= 539484.3 J
b) Power = work done /time
= 539484.3 / 7.5
= 71931.24 Watt
c) heat generated = work done by friction = -(change in K.E) ( Vf=0 Vi = 55 miler/hr
= 0.5 x m x (Vi ^2 - Vf ^2) => Vi = 55 x 1609.34 m/3600s
= 0.5 x 1500 x (24.587 ^2 ) = 24.587 m/s
= 453390.43 J
= 108.31 Kcal
Qsn 8) Kinematics: from v^2 - u^2 = 2as where v=0 (at max height) u = 350 m/s a= -g = -9.8 m/s^2 s=h
=> 0 - (350^2) = 2 x (-9.8) x h
=> h = 6250 m
Energy : as the only force is gravity and it is conservative total energy of the system remains same.
T.E = P.E. + K.E.
initially P.E = 0 K.E = 0.5 x m x V^2
finally P.E = m x g x h K.E = 0
=> 0 + 0.5 x m x V^2 = m x g x h + 0
=> h = V^2 / (2x g)
= 350 ^2 / ( 2 x 9.8)
= 6250 m
Qsn 6) A) P.E in spring = K.E of bullet
0.5 x K x (x^2) = 0.5 x m x V^2
=> 0.5 x K x (0.04 ^2) = 0.5 x 0.01 x 5^2
=> K = 156.25 N/m
B ) if v = 7 m/s
K.E = 0.5 x 0.01 x 7^2
P.E = 0.5 x 156.25 x x^2
=> 0.5 x 156.25 x x^2 = 0.5 x 0.01 x 7^2
=> x = 5.6 cm
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