Question

A block rests on a horizontal, frictionless surface. A string is attached to the block, and is pulled with a force of 48.0 N at an angle θ above the horizontal. After the block is pulled through a distance of 16.0 m its speed is v = 2.10 m/s, and 40.0 J of work has been done on it.What is the mass of the block? (Answer in kg)

Answer 1

From the given that first we need find the angle then

Work done(W) =Force *dispalcement

W =Fcostheta*s

40 =48costheta*16

then theta =Cos-1(40/16*48) =87.01degrees

Time taken to move 16 m is given by

V =u+at     intial speed (u) =0 and v =2.10m/s

V =0+at

t =v/a =2.10/a

Now from the relation s =ut+(1/2)at²

s =0+0.5at²

16 =0.5a*(2.10/a)2 ====>a =0.5*(2.10)²/16 =0.137m/s²

Now from relation

ma =48costheta =48*cos(87.01)

m =48*cos(87.01)/0.137=18.275kg