Answer:
The correct answer is:
(a) 0
(b) 0.5 m/s
(c) 7740 N
(d) 0
Step-by-step 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* 9.8)-(7200* 0.5)]](https://img.qammunity.org/2022/formulas/physics/college/mf7asbqd6wq1qjd7jlz7h3l3rvzpt3m2xu.png)
=
![0.3[29400-3600]](https://img.qammunity.org/2022/formulas/physics/college/85g4aqmrspnkroyag859gr9fsqtb136ik4.png)
=

=

(d)
On including the friction,
The net work will be:
⇒
