Answer:
The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.
Explanation:
It is given that,
Mass of the box, m = 2.2 kg
The box is inclined at an angle of 30 degrees
Vertical distance, d = 3.1 m
The coefficient of friction, 
Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.
W is the work done by the friction.
v = 8.19 m/s
So, the speed of the box is 8.19 m/s. Hence, this is the required solution.
Answer:
0.456033049
Explanation:
where N=mg hence
where m is mass of object, g is acceleration due to gravity whose value is taken as 
, 
is the coefficient of static friction and F is the applied force.
Making the subject we obtain
and substituting m for 38 Kg, g for and 170 N for F we obtain
Therefore, the coefficient of static friction is 0.456033049
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>
m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... y direction
Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>
Net force = √625 = 25N
F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2
Answer: 5.00 m/s^2
b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.
</span>
<span>m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... 60 degress above x direction
Components of F2
F2,x = F2*cos(60) = 15N / 2 = 7.5N
F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N
Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N
Total force in y = F2,y = 13.0 N
Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =
= 30.42N
a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2
Answer: 6.08 m/s^2
</span>
The correct answer is <span>3)
.
</span>
In fact, the total energy of the rock when it <span>leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
</span>
<span>As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy:
</span>
<span>
But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
</span>
<span>
</span>
The force exerted on the car during this stop is 6975N
<u>Explanation:</u>
Given-
Mass, m = 930kg
Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s
Time, t = 2s
Force, F = ?
F = m X a
F = m X s/t
F = 930 X 15/2
F = 6975N
Therefore, the force exerted on the car during this stop is 6975N
