1 ) A 2500 kg car accelerates up to a velocity of 30.0 m/s in 8.00 seconds. The

Question

1 ) A 2500 kg car accelerates up to a velocity of 30.0 m/s in 8.00 seconds. The work done by the engine to accelerate the car is
A) 3.10 x 10^6 J.
B) 2.56 x 10^6 J.
C) 2.01 x 10^6 J.
D) 1.85 x 10^6 J.
E) 1.13 x 10^6 J.

2) A 2.00 kg pendulum bob on a string 1.50 m long, is released with a velocity of 2.00 m/s when the support string makes an angle of 30.0 degrees with the vertical. What is the velocity of the bob at the bottom of the swing?
A) 4.32 m/s
B) 4.00 m/s
C) 3.75 m/s
D) 3.04 m/s
E) 2.81 m/s

Explanation / Answer

1.
Acceleration = a = (30-0)/8 = 3.75 m/s^2

Now,
v^2 = u^2 + 2as
30^2 = 0 + 2 x 3.75 x s
Distance = s= 120 meters
Work = F.s = 2500 x 3.75 x 120 = 1.13 x 10^6 Joules

Option- E is correct !!!

2.
we have,
0.5mv^2 - 0.5mu^2 = mgL(1-cos)

Speed = v = 2.82 m/s

Option - E is correct !!

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