1. A student pulls horizontally on a 12 kg box, which then moves horizontally wi
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Question
1. A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s2. If the student uses a force of 15 N, what is the coefficient of kinetic friction of the floor?
2. A 30 kg block slides down a 20° ramp with an acceleration of 1.2 m/s2. What is the coefficient of kinetic friction between the block and the ramp?
3. For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at = 45°. Substantiate your answer with mathematical support.
4. A stone is projected horizontally at a speed of 6 m/s form the top of a 19.6 m building. What is the final velocity with which the stone would hit the ground?
5. A man swims across the river with a velocity of 2 m/s relative to the flow of water. If the river flows steadily at 4 m/s toward south, what is the velocity of the man with respect to the shore? Solve for both the magnitude and the direction (in degrees).
2. A 30 kg block slides down a 20° ramp with an acceleration of 1.2 m/s2. What is the coefficient of kinetic friction between the block and the ramp?
3. For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at = 45°. Substantiate your answer with mathematical support.
4. A stone is projected horizontally at a speed of 6 m/s form the top of a 19.6 m building. What is the final velocity with which the stone would hit the ground?
5. A man swims across the river with a velocity of 2 m/s relative to the flow of water. If the river flows steadily at 4 m/s toward south, what is the velocity of the man with respect to the shore? Solve for both the magnitude and the direction (in degrees).
Explanation / Answer
First, find the weight of the box: W = mg.
m = 12 kg; g = 9.8 m/s², so W = 118 N. This, times the coefficient of friction k, gives the frictional force opposing the motion: Wk, or 118k N.
Now use F = ma to find the force needed to accelerate the box, if it were moving without friction:
m = 12 kg; a = 0.2 m/s², so F = 2.4 N.
The applied force is 15 N, so the part of the force that is needed to overcome friction is 12.6 N.
12.6 N = Wk = (118 N)(k), so k = 12.6/118 = 0.11 (to 2 significant figures).
2)
The total force causing the acceleration is composed of two parts:
the component of gravity downwards along the ramp: m g sin(20 degrees)
the friction force upward along the ramp: F_friction = F_normal = m g cos(20 degrees)
So Newton's second law gives, for the sliding acceleration:
m a = m g sin(20 degrees) - m g cos(20 degrees)
The mass cancels and we are left with
= (g sin(20) - a)/(g cos(20) )
= (9.81 m/s² * 0.342 - 1.2 m/s²)/( 9.81 m/s² * 0.940)
= 0.23
4) 0.5 (m)(6)^2 + m(19.6)(9.81) = 0.5 m (v)^2
m's cancel out
18 + 192.276= 0.5 v^2
V^2= 420.552
v= 20.5
5)
velocity of 2 m/s relative to the flow of water at 4 m/s toward south.
V = (Vacross^2 +Vwater^2) = (2^2 +4^2) = (4+16)
How far does the box move? They want meters not sec
What is the final velocity? in m/s
mv^2/2 = mgh + m(v0)^2/2
v = (2gh + (v0)^2)
v = (2*9.8*19.6 +6.0^2) =20.5 m/s
3)
The maximum horizontal distance is possible only at = 45°.
Maximize x, when y goes from 0 to max and back to 0
x= v t cos
y = 0 = v t sin - 1/2 g t^2
v sin = 1/2 g t
t =2 v sin/g
x= v 2 ( v sin/g) cos
x= v^2 (2 sin cos)/g
x= v^2 sin(2)/g which is max when 45° (sin 45°=1)
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