A block of mass 2 kg is initially at rest on a horizontal surface. At time , we begin pushing on it with a horizontal force that varies with time as , where 0.6 N/s . We stop pushing at time s [ for ].
(a) First, assume the surface is frictionless. What is the magnitude of the final momentum of the block at s? (in kg m/s)
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(b) Let us now consider a new situation where the object is initially at rest on a rough surface. The coefficient of static friction is . What is the speed of the block at time s?. For simplicity, we take static and kinetic friction coefficients to be the same, and consider m/s .
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(c) What is the power provided by the force at time s (in Watts) in the case where there is friction (part (b)) ?
The first part is easy: change in momentum is impulse, which you can find by integrating F from 0 to your time.
Now, what about b)?
The same goes for part b), but when you integrate you should start from the time the mass moves.
SO change in momentum equal change in impulse only from the time they're moving
you find speed and c) is trivial
P = F v, both at your final time
have you got the answer for b & c?
*The same goes for part b), *
don't you have to integrate over the resulting force, that is (beta*t^2-friction)?
I still can't understand how to integrate... would you show me?
To answer part (a), we can use the concept of impulse-momentum theorem. The impulse of a force is defined as the change in momentum it imparts to an object. The impulse is equal to the force multiplied by the time over which it acts. In this case, the force varies with time, so we need to determine the impulse over the given time interval.
The impulse can be calculated using the definite integral of the force function over the time interval. The force function is given as F(t) = 0.6t N, where t is in seconds.
To obtain the impulse, we integrate the force function from t = 0 to t = 6 seconds:
Impulse = ∫(0.6t) dt, limits from 0 to 6.
Integrating the force function gives:
Impulse = [0.3t^2] from 0 to 6.
Substituting the limits:
Impulse = 0.3(6)^2 - 0.3(0)^2.
Impulse = 0.3(36) - 0.3(0).
Impulse = 10.8 Ns (or kg m/s).
Therefore, the magnitude of the final momentum of the block at 6 seconds is 10.8 kg m/s.
Now moving on to part (b), where there is a rough surface and static friction involved.
The force of static friction can be calculated using the equation:
Frictional Force = coefficient of static friction * normal force.
The normal force is equal to the weight of the block, which is given as mass * acceleration due to gravity (m * g).
The coefficient of static friction is not provided, so we cannot calculate the exact value. However, we can determine the maximum possible static frictional force using the equation:
Maximum Frictional Force = coefficient of static friction * normal force.
Substituting the given values, we have:
Maximum Frictional Force = coefficient of static friction * (2 kg * 9.8 m/s^2).
The block is initially at rest, so the static frictional force must be equal and opposite to the pushing force at any given time. So we can equate the maximum frictional force to the push force function:
coefficient of static friction * (2 kg * 9.8 m/s^2) = 0.6t N.
Solving for the coefficient of static friction:
coefficient of static friction = (0.6t) / (2 * 9.8).
Now let's move on to part (c) to calculate the power provided by the force at time t.
Power is defined as the rate at which work is done or energy is transferred. In this case, the force is the external force exerted on the block.
The power can be calculated using the equation:
Power = force * velocity.
At any given time t, the velocity of the block can be determined by integrating the acceleration function of the block over the time interval from 0 to t. However, as the acceleration function is not provided, we cannot calculate the exact value. Therefore, for simplicity, let's assume a constant acceleration of 5 m/s^2, as mentioned in the problem.
Using the equation of motion:
velocity = initial velocity + acceleration * time.
The initial velocity is 0 m/s, and the acceleration is 5 m/s^2. Substituting the values, we have:
velocity = 0 + 5t.
Now we can calculate the power at time t:
Power = force * velocity.
Using the force function given in the problem (0.6t N) and the velocity we just derived (5t m/s):
Power = (0.6t) * (5t).
Simplifying:
Power = 3t^2 watts.