Hey can you plz help me
1 answer:
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Answer:
(a) 0.38 m
(b) 2.78 m/s
(c) 0.11 watt
Explanation:
mass, m = 0.3 kg
spring constant, K = 160 N/m
initial compression, d = 12 cm = 01.2 m
initial speed, u = 3 m/s
(a) Let the maximum stretch is y.
Use conservation of energy
Initial potential energy + initial kinetic energy = final potential energy
0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²
160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²
2.304 + 0.00432 = 160 y²
y = 0.38 m
y = 38 cm
(b) Let v is the maximum speed.
The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy
Initial potential energy + initial kinetic energy = final kinetic energy
0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²
160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²
2.304 + 0.00432 = 0.3 v²
v = 2.78 m/s
(c) The time period of the spring mass system is given by
T = 0.272 second
Energy dissipated per cycle = 0.03 J
Power, P = 0.03 / 0.272 = 0.11 Watt
Answer:
The value of tension on the cable T = 1065.6 N
Explanation:
Mass = 888 kg
Initial velocity ( u )= 0.8
Final velocity ( V ) = 0
Distance traveled before come to rest = 0.2667 m
Now use third law of motion =
- 2 a s
Put all the values in above formula we get,
⇒ 0 = - 2 × a ×0.2667
⇒ a = 1.2
This is the deceleration of the box.
Tension in the cable is given by T = F = m × a
Put all the values in above formula we get,
T = 888 × 1.2
T = 1065.6 N
This is the value of tension on the cable.