Question

A block of mass 5 kg is initially lying at rest on a smooth floor. My friend exerts a constant force of 50 N on it. If I exert a force of T on the block, opposing my friend’s force, the block has an acceleration of magnitude aa. If I double my force to 2T, the magnitude of block’s acceleration remains the same. Find the magnitude of the acceleration of the block when I triple my force to 3T.

Answer 1

Explanation:

Using work-energy theorem:

W.D

force

+W.D

friction

=ΔK.E

45×2−25×2=KE

final

KE

final

=40 J

solution

Answer 2

Answer:

10 ms^-2

Explanation:
F = ma

case 1

F= 50 -T

m=5kg

50 - T = 5a .............(1)

case 2

F = 2T -50
2T -50 =5a..............(2)

Solve (1) and (2)

T=10a

a =10 / 3

T = 100/3 N

Therefore 3T = 100 N

Please mark as brainliest.