A block of mass m= 3 kg is initially at rest on a horizontal surface. At time t=0, we begin pushing on it with a horizontal force that varies with time as F(t)=βt2, where β= 1.2 N/s2. We stop pushing at time t1=5 s [F(t)=0 for t>t1].
(a) First, assume the surface is frictionless. What is the magnitude of the final momentum of the block at t1=5 s? (in kg m/s)
pfin(t=t1)=
(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 μs=0.2. What is the speed of the block at time t2=5 s?. For simplicity, we take static and kinetic friction coefficients to be the same, μs=μk and consider g=10 m/s2.
v(t=t2)=
(c) What is the power P provided by the force F(t) at time t3=4 s (in Watts) in the case where there is friction (part (b)) ?
P(t=t3)=
(a) In the case of a frictionless surface, the net force acting on the block is the applied force.
The formula for force is F(t) = βt^2
To find the final momentum at t1 = 5 s, we need to integrate the force function over the time interval from 0 to t1.
Momentum (p) is given by the formula:
p = ∫ F(t) dt
p = ∫ βt^2 dt (from 0 to t1)
Integrating βt^2 gives us (β/3) * t^3.
Plugging in t1 = 5 s, we have:
pfin(t=t1) = (β/3) * t1^3
pfin(t=t1) = (1.2/3) * (5)^3
pfin(t=t1) = 20 kg m/s
Therefore, the magnitude of the final momentum of the block at t1 = 5 s is 20 kg m/s.
(b) In the case where there is friction, we need to consider the force of friction opposing the motion of the block.
The force of friction is given by the equation:
Ffriction = μ * m * g
where μ is the coefficient of friction, m is the mass of the block, and g is the acceleration due to gravity.
In this case, μ = μs = μk = 0.2 and g = 10 m/s^2.
The net force acting on the block is then:
Net force = F(t) - Ffriction
At t = t2 = 5 s, the velocity of the block is the integral of the net force function over time from 0 to t2.
Velocity (v) is given by the formula:
v = ∫ (F(t) - Ffriction) dt
v = ∫ (βt^2 - μ * m * g) dt (from 0 to t2)
Integrating βt^2 gives us (β/3) * t^3.
Plugging in t2 = 5 s and the given values of μ and g, we have:
v(t=t2) = ∫ [(β/3) * t^3 - μ * m * g] dt
v(t=t2) = ∫ [(1.2/3) * t^3 - 0.2 * 3 * 10] dt (simplifying the expression)
v(t=t2) = [(1.2/3) * t^4 - 6] (from 0 to t2)
v(t=t2) = [(1.2/3) * (5)^4 - 6]
v(t=t2) ≈ 23 m/s
Therefore, the speed of the block at t2 = 5 s is approximately 23 m/s.
(c) Power (P) is defined as the rate at which work is done. It can be calculated as the derivative of work with respect to time.
The formula for power is:
P = dW/dt
Since work (W) is equal to force times distance, we can write:
P = F * v
At t = t3 = 4 s, the power provided by the applied force is given by:
P(t=t3) = F(t3) * v(t=t3)
We can find F(t3) by plugging t3 = 4 s into the force equation:
F(t=t3) = β * t3^2
F(t=t3) = 1.2 * (4)^2
F(t=t3) = 19.2 N
Substituting this value and the velocity v(t=t3) from part (b) into the power equation, we have:
P(t=t3) = 19.2 * 23
P(t=t3) = 441.6 Watts
Therefore, the power provided by the force F(t) at t3 = 4 s is 441.6 Watts.
(a) To find the magnitude of the final momentum of the block at t1=5s, we need to integrate the force over time to find the change in momentum.
The force acting on the block is given by F(t) = βt^2 = 1.2t^2 N.
Since the force is changing over time, we need to calculate the impulse by integrating the force over time from t=0 to t=5s:
Impulse = ∫ F(t) dt = ∫ 1.2t^2 dt
Integrating 1.2t^2 gives us (1.2/3)t^3.
Evaluating the integral from 0 to 5, we have:
Impulse = (1.2/3)(5^3 - 0) = (1.2/3)(125 - 0) = 50 Ns
The change in momentum is equal to the impulse, so the magnitude of the final momentum of the block is 50 kg m/s.
Therefore, pfin(t=t1) = 50 kg m/s.
(b) In this second scenario, where the object is initially at rest on a rough surface, we need to consider the effect of friction. The block will experience a frictional force opposing its motion.
The frictional force can be calculated using the static friction coefficient (μs) and the normal force (N). The normal force can be determined as the weight of the block, which is equal to m * g, where g is the acceleration due to gravity (10 m/s^2 in this case).
The maximum static frictional force can be found using the equation: fs ≤ μs * N
fs ≤ (0.2) * (m * g)
The block will start moving once the force applied exceeds the maximum static frictional force. Since we already know the applied force F(t) is 1.2t^2 N, we can find the time t2 at which the block starts moving by setting F(t2) equal to the maximum static frictional force:
1.2t2^2 = (0.2) * (m * g)
1.2t2^2 = (0.2) * (3 kg * 10 m/s^2)
1.2t2^2 = 6 N
t2^2 = 6 / 1.2
t2^2 = 5
t2 = √5
Now, we can find the speed of the block at time t2=5s using the equation for speed, v = u + at, where u is the initial velocity (0 m/s), a is the acceleration, and t is the time:
v(t=t2) = 0 + (1.2/3) * t2^2
v(t=t2) = 0 + (1.2/3) * 5^2
v(t=t2) = 0 + (1.2/3) * 25
v(t=t2) = 0 + 10 m/s
Therefore, the speed of the block at time t2=5s is 10 m/s.
(c) To find the power P provided by the force F(t) at time t3=4s, we need to calculate the rate at which work is done by the force. Power is defined as P = dW / dt, where dW is the differential work done and dt is the differential time.
The work done by a force is given by the equation: W = ∫ F(t) ds, where ds is the differential displacement. Since the force is applied horizontally, the displacement and the force are in the same direction.
Assuming the block moves with constant speed after it starts moving, we can calculate the work done using the equation:
W = ∫ F(t) ds = ∫ F(t) v(t) dt
W = ∫ (1.2t^2) v(t) dt
To find the power at t3=4s, we need to evaluate the integral from t=0 to t=4:
P(t=t3) = dW / dt = ∫ (1.2t^2) v(t) dt from 0 to 4
To evaluate this integral, we need to know the function v(t). Since we do not have this information, we cannot calculate the power provided by the force at time t3=4s without additional information.