In fig. let the mass of the block be 8.5 kg and the angle 9 be 30°. (g = 10 m/s?).
smoothm
0
Tension in the cord
Answers
Explanation:
ANSWER
The free-body diagram of the problem is shown to the right.
Since the acceleration of the block is zero, the components of the Newton’s second law equation yield
T–mgsinθ=0
F
N
−mgcosθ=0,
where T is the tension in the cord, and N is the normal force on the block.
(a) Solving the first equation for the tension in the string, we find
T=mgsinθ(85kg)(9.8m/s
2
)sin30
o
=42N
(b) We solve the second equation in part (a) for the normal force F
N
.
F
N
=mgcosθ(85kg)(9.8m/s
2
)sin30
o
=72N
(c) When the cord is cut, it no longer exerts a force on the block and the block accelerates.
The x-component equation of Newton’s second law becomes −mgsinθ=ma, so the acceleration becomes
a=−gsinθ=−(9.8m/s
2
)sin30
o
=−4.9m/s
2
The negative sign indicates the acceleration is down the plane.
The magnitude of the acceleration is $$4.9 \, m/s^2$$.