Help please! Can someone show me how to solve this and the answers too? . A ball

Question

Help please! Can someone show me how to solve this and the answers too? . A ball of mass 0.7 kg is pressed against compressed spring and then the ball and sprin are released from rest. The spring applies a constant force of 0.25 Newtons over a distance of 3 meters. The ball then proceeds to roll along a frictionless surface until it encounters a ramp which rises 25 degrees above the horizontal and has a coefficient of rolling friction of = 0.05. (A) What is the acceleration of the ball while the spring is pushing it? (5 pts) (B) What is the speed of the ball immediately before it rolls onto the ramp? (5 pts) (C) How high above the horizontal does the ball roll? (15 pts) 0.3 meters = 0.05 25 25° Frictionless Surface

Explanation / Answer

Here the spring applies a constant force of 0.25 Newtons over a distance of 0.3 metres. We know that for a force F acting on a body over a distance of s, the work done is given as: W = F.S. We will make use of this to determine the energy imparted to the ball by the action of the spring and hence find the values to be determined as follows:

A.) As the spring applies a constant force, the acceleration would be given as: F/m

or acceleration = 0.25 / 0.7 = 0.357 m/s2 is the required acceleration of the ball.

B.) Once the ball comes off the spring, it will have some kinetic energy for the speed it will attain by the action of the force due to the spring. By principle of conservation of energy we can say that the kinetic energy of the ball will be equal to the work done by the spring. We can write an equation involving the kinetic energy of the ball and the work done:

That is, F.x = 0.5 mv^2

or, 0.25 (0.3) = 0.5 (0.7) v^2

or, V = 0.6804 m/s is the velocity of the ball with which it leaves the spring.

Therefore the speed of the ball just before it gets on the ramp is 0.6804 m/s

C.) While the ball starts moving on the ramp, the friction will do negative work on the ball along with the gravity, as a result of which the ball will come to a halt at the topmost point it would reach.

We will use the principle of conservation of energy to form an equation involving the height it reaches, work done by the friction and the initial kinetic energy. Also, we will use the bottom horizontal surface as the reference for calculation the potential energy.

Hence we ca write: 0.075 = mgH + (mgCos25)r(H/sin25)

or, H (0.7*9.81 + 0.7*9.81*cos25*0.05/sin25)= 0.075

or H(7.60332) = 0.075

or H = 0.009864 metres is the required height upto which the ball rises.

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