You attach a 2.90 kg mass to a horizontal spring that is fixed at one end. You pull the mass until the spring is stretched by 0.
1 answer:
Answer:
Explanation:
The spring is stretched by .5 m and then released that means its amplitude of oscillation A is 0.5 m .
A = 0.5 m
After the release at one extreme point , the mass comes to rest again at another extreme point after half the time period ie
T / 2 = .3 s
T = 0.6 s
Angular velocity
ω =
ω =
ω = 10.45
Maximum velocity = ω A
ω and A are angular velocity and amplitude of oscillation.
Maximum velocity = 10.45 x .5
= 5.23 m /s
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Answer:
False
Explanation:
The tension in the chain must be equal to the frictional force acting on the car, not to its weight.
In fact, we have 4 forces acting on the car
- Its weight: downward
- The normal reaction of the road on the car: upward --> this force balances the weight, so the net force along the vertical direction is zero
- The tension in the chain: forward
- The frictional force between the road's surface and the tires of the car: backward
We can consider the horizontal motion only: we are said that the car is moving at constant velocity, so the horizontal acceleration is zero. According to Newton's second law:
zero acceleration means that the resultant of the forces on the car is zero. But there are only 2 forces acting on the car in the horizontal direction: the tension in the chain (forward) and the frictional force (backward). Since their resultant must be zero, it means that the two forces must be equal and opposite: therefore, the tension in the chain must be equal to the frictional force.