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:
⇒ [tex]\Delta K =0[/tex]
By using the theorem of energy, the net work done will be:
⇒ [tex]\Delta K =0[/tex]
(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:
= [tex]u(mg-Tsin30^{\circ})[/tex]
On substituting the values, we get,
= [tex]0.3[(3000\times 9.8)-(7200\times 0.5)][/tex]
= [tex]0.3[29400-3600][/tex]
= [tex]0.3\times 25800[/tex]
= [tex]7,740 \ N[/tex]
(d)
On including the friction,
The net work will be:
⇒ [tex]\Delta K=0[/tex]
