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 ө = 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).

Answer 1

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 μk, gives the frictional force opposing the motion: Wμk, or 118μk 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 = Wμk = (118 N)(μk), so μk = 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 = μ F_normal = μ m g cos(20 degrees)

So Newton's second law gives, for the sliding acceleration:

m a = m g sin(20 degrees) - μ m g cos(20 degrees)

The mass cancels and we are left with

μ = (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 ө = 45°.
Maximize x, when y goes from 0 to max and back to 0
x= v t cosθ
y = 0 = v t sinθ - 1/2 g t^2
v sinθ = 1/2 g t
t =2 v sinθ/g
x= v 2 ( v sinθ/g) cosθ
x= v^2 (2 sinθ cosθ)/g
x= v^2 sin(2θ)/g which is max when θ ισ 45° (sin 45°=1)