Question

A force F=57√2
t N is applied on the block as shown in
the figure. The block starts from rest at t = 0, then
(a) the block leaves the horizontal surface at t = 20 s
(b) the block never leaves the horizontal surface
(c) the velocity of block at t = 20 s is 100 m/s
(d) Both (a) and (c) are correct​

Answer 1

Answer:

Given,

m=10kg

Initially block is at rest, so initial momentum is zero.

From the free body diagram,

N=mg=5

2

tsin45

0

But only horizontal force is responsible for the motion of block that is,

F=5

2

tcos45

0

=5t

Newton's second law of motion

F=

dt

dP

dP=F.dt

P=∫

0

20

5tdt

mv=1000kgm/s

v=100m