A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
1 answer:
Answer:
a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.
b) The final speed of the block of ice is 9.8 meters per second.
Explanation:
a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.
The weight of the block (), measured in newtons, is:
(1)
Where:
- Mass of the block of ice, measured in kilograms.
- Gravitational acceleration, measured in meters per square second.
If we know that and
, the magnitudes of the weight and normal force of the block of ice are, respectively:
And the pull force is:
The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.
b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:
(2)
Where:
,
- Initial and final speeds of the block, measured in meters per second.
- Horizontal net force, measured in newtons.
- Impact time, measured in seconds.
Now we clear the final speed in (2):
If we know that ,
,
and
, then final speed of the ice block is:
The final speed of the block of ice is 9.8 meters per second.
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Answer:
a) Ffr = -0.18 N
b) a= -1.64 m/s2
c) t = 9.2 s
d) x = 68.7 m.
e) W= -12.4 J
f) Pavg = -1.35 W
g) Pinst = -0.72 W
Explanation:
a)
- While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
- This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
- So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.
where μk is the kinetic friction coefficient, and Fn is the normal force.
- Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and the weight Fg, downward), we conclude that both must be equal and opposite each other:
- We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:
b)
- According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
- In this case, this net force is the friction force which we have already found in a).
- Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
- We can write the expression for a as follows:
c)
- Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:
- Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:
d)
- From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
- So, we can use the following kinematic equation in order to find the displacement before coming to rest:
- Since the puck comes to a stop, vf =0.
- Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:
e)
- The total work done by the friction force on the object , can be obtained in several ways.
- One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
- Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:
f)
- By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
- If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:
g)
- The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
- Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows: