Solve these numericals:- (PLEASE HELP, URGENTLY NEEDED!)
1) A certain force exerted for 1.2 seconds raises the speed of an object from 1.8 metres per second to 4.2 metres per second. Later the same force is applied for 2 seconds. How much does the speed change in two seconds?
2) A body is released from a height of 20 metres. Calculate the final velocity of the body (take g = 10 metres per second square).
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1) Force = F so acceleration a = F/m
a = (v - u) / t = (4.2 - 1.8) / 1.2 = 2 m/sec²
since same force is applied again, the same acceleration happens.
v = u + a t => v - u = a t = 2 m/sec² * 2sec = 4 m/sec
change in speed = 4 m/s
2) s = 20 meters. = displacement
v² = u² + 2 a s -- equation of motion under gravity, a = + g
v² = 0 + 2 g s = 2 * 10 * 20
v = 20 m/sec = final velocity
a = (v - u) / t = (4.2 - 1.8) / 1.2 = 2 m/sec²
since same force is applied again, the same acceleration happens.
v = u + a t => v - u = a t = 2 m/sec² * 2sec = 4 m/sec
change in speed = 4 m/s
2) s = 20 meters. = displacement
v² = u² + 2 a s -- equation of motion under gravity, a = + g
v² = 0 + 2 g s = 2 * 10 * 20
v = 20 m/sec = final velocity
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