Answer:
a. N = 2.49W b. 0.40
Explanation:
a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?
Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s
Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m
The rider's weight W = mg = 9.8m
The ratio of the normal force to the rider's weight is
N/W = 24.43m/9.8m = 2.49
So the normal force expressed in term's of the rider's weight is
N = 2.49W
b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?
The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.
Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.
So, the normal force equals
N = F/μ = W/μ = mg/μ = mrω²
μ = mg/mrω²
= W/N
= 9.8m/24.43m
= 0.40
Answer:
The increase in the internal energy = 350 J
Explanation:
Given that
Q= 275 J
W= - 125 J
W' = 50 J
W(net)= -125 + 50 = -75 J
Sign -
1.Heat rejected by system - negative
2.Heat gain by system - Positive
3.Work done by system = Positive
4.Work done on the system-Negative
Lets take change in the internal energy =ΔU
We know that
Q= ΔU + W(net)
275 = ΔU -75
ΔU= 275 + 75 J
ΔU=350 J
The increase in the internal energy = 350 J
Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.
Answer:
Friction acts in the opposite direction to the motion of the truck and box.
Explanation:
Let's first review the problem.
A moving truck applies the brakes, and a box on it does not slip.
Now when the truck is applying brakes, only it itself is being slowed down. Since the box is slowing down with the truck, we can conclude that it is friction that slows it down.
The box in the question tries to maintains its velocity forward when the brakes are applied. We can think of this as the box exerting a positive force relative to the truck when the brakes are applied. When we imagine this, we can also figure out where the static friction will act to stop this positive force. Friction will act in the negative direction. Or in other words, friction will act in the opposite direction to the motion of the truck and box. This explains why the box slows down with the truck, as friction acts to stop its motion.
As per given equation we have
now as per the dimensional analysis we can say that dimension of right side of equation must be equal to left side of the equation
now as per left side of equation its dimension is same as length or meter
now we can say it should be meter on right side also
similarly for other term we have
<em>so above are the dimensions of b and c</em>
