A student holds one end of a string in a fixed position. A ball of mass .2 kg attached to the other end of the string moves in a horizontal circle of radius .5m with a constant speed of 5m/s. How much work is done on the ball by the string during each revolution?
a)0 J
b).5 J
c)1 J
d)2pi J
e)5pi J
Could you please tell me how I would solve this problem? I used F=mv^2/r and then I got 10. Then I used W=fd which gave me W=(10)(.5) and I got 5 as the answer. But this answer isn't in the choices.
The answer is 0 j.
Work done equals displacement times the component of the force in the direction of the displacement. In this case the force exerted by the string on the ball is always at right angles to the direction in which the ball moves.
You can also say that the work done must equal the change in kinetic energy of the ball, but the ball moves at constant speed, only the direction of the velocity changes. So, the kinetic energy stays constant and therefore no work is done on the bal.
Or, you can think of it as you start at an A point on the circle, after one revolution, it returns back to A, means displacement is 0. Since W=Displacement*Force so W=0
It's in a uniform circular motion so no work is done.
Well, it seems like the string is just stringing you along with this question! The poor ball is going around in circles, but it's not doing any work because the force from the string is always perpendicular to the direction of motion. So, the answer is 0 J, which is about as exciting as watching a circus with no clowns! But hey, at least the ball gets to have a free ride without any work to do. Lucky ball!
To solve this problem, let's start by finding the tension in the string. The tension in the string is equal to the centripetal force required to keep the ball moving in a circle.
The formula for centripetal force is given by F = (m*v^2)/r, where m is the mass of the ball, v is the velocity of the ball, and r is the radius of the circle.
Plugging in the given values, we have F = (0.2 kg * (5 m/s)^2) / 0.5 m = 10 N.
Now, to find the work done on the ball by the string during each revolution, we need to calculate the work done by the tension force.
Work done is given by the formula W = F*d*cosθ, where F is the force, d is the displacement, and θ is the angle between the force and the displacement.
In this case, the displacement d is the circumference of the circle, which is 2πr, and the angle θ between the force and the displacement is 90 degrees since the force is always perpendicular to the displacement.
Therefore, the work done is W = (10 N) * (2π * 0.5 m) * cos(90 degrees).
However, cos(90 degrees) is equal to 0, which means the work done is 0 J.
So, the correct answer is option a) 0 J.
To solve this problem, we need to understand that work is done when a force is applied in the direction of displacement. In this case, the ball is moving in a horizontal circle, and the force exerted by the string on the ball is always perpendicular to the direction of motion. This means that the string is not doing any work on the ball.
One way to see this is by considering the formula for work: W = F*d*cos(theta), where F is the force, d is the displacement, and theta is the angle between the force and displacement vectors. Since the force and displacement are always at right angles to each other, the angle theta is 90 degrees, and cos(theta) becomes 0. Therefore, the work done on the ball is 0 J.
Another way to see this is by considering the conservation of energy. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the ball is moving at a constant speed, which means its kinetic energy is constant. Therefore, there is no change in kinetic energy, and no work is done on the ball.
In conclusion, the correct answer to this problem is a) 0 J, as no work is done on the ball by the string during each revolution.