Answer:
The correct answer is:
(a) 0
(b) 0.5 m/s
(c) 7740 N
(d) 0
Explanation:
The given values are:
mass,
m = 3000 kg
Tension,
T = 7,200 N
Angle,
= 30°
(a)
Even as the block speed becomes unchanged, the kinetic energy (KE) will adjust as well:
⇒ 
By using the theorem of energy, the net work done will be:
⇒ 
(b)
According to the question, After 0.25 m the block is moving with the constant speed
= 0.5 m/s.
(c)
The given kinetic friction coefficient is:
u = 0.3
The friction force will be:
=
On substituting the values, we get,
= ![0.3[(3000\times 9.8)-(7200\times 0.5)]](https://tex.z-dn.net/?f=0.3%5B%283000%5Ctimes%209.8%29-%287200%5Ctimes%200.5%29%5D)
= ![0.3[29400-3600]](https://tex.z-dn.net/?f=0.3%5B29400-3600%5D)
= 
= 
(d)
On including the friction,
The net work will be:
⇒ 