A block having a mass of 3 kg slides on a frictionless floor. At the instant tha

Question

A block having a mass of 3 kg slides on a frictionless floor. At the instant that time t = 0, the block's velocity is [ 2i - 5j ] m/sec; at t = 0, the block is acted upon by a constant force of [ - 120i + 240j ] N. Assuming that the force acts on the block during the time interval 0 lessthanorequalto t lessthanorequalto 0.10 sec, what is the block's velocity at the instant t = 0.10 sec ? (in m/sec) 4i - 4j -6i + 13j -2i + 3j 16i - 24j 6i - 13 j We are told that a tennis ball (mass - M) has a velocity of magnitude v in the -x direction at the instant that it is struck by a tennis racket; immediately after being struck, the ball has a velocity of 4v in the +x direction. If the ball was in contact with the racket for 0.001 sec, what was the magnitude and direction of the average force exerted on it by the racket? 5000Mv in the +x direction 4000Mv in the +x direction 3000Mv in the +x direction 3000Mv in the -x direction 5000Mv in the -x direction

Explanation / Answer

since the force is constant , the acceleration of the body
a = F / M
a = -120i^ + 240 j^ / 3 kg
a = -40 i^  + 80 j^
The initial velocity is
u = 2 i^ - 5 j ^
In the x direction the final velocity is
vx  = ux  + ax t
vx  =   2 - 40 x 0.10
vx  = - 2
The y compnent of the final velocity
vy  = uy + ay t
vy   = - 5 + 80 x 0.10
vy   = 3
Thus the final velocity in the vector form
v = - 2 i^  + 3 j^

Average force F = M (vf  - vi ) / t
Initial velocity vi  = - v , final velocity vf  = 4 v
F = M ( 4 v - - v) / 0.001s
F = 5 Mv / 0.001
F = 5000 Mv
In the + x direction

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