Answer:
Explanation:
Potential Energy= Kinetic Energy
Let be the value of Kinetic Energy.
We know that
Make the subject of the formula to get speed at the bottom of the hill.
A 2kg mass moving at a speed of 3 m/s is stopped by a constant force of 15N. How many seconds must the force act on the mass to
1 answer:
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Answer:
a) u_s=0.375
b )v=14.4 m/s
Explanation:
1 Concepts and Principles
Particle in Equilibrium: If a particle maintains a constant velocity (so that a = 0), which could include a velocity of zero, the forces on the particle balance and Newton's second law reduces to:
∑F=0 (1)
2- Particle in Uniform Circular Motion:
If a particle moves in a circle or a circular arc of radius Rat constant speed v, the particle is said to be in uniform circular motion. It then experiences a net centripetal force F and a centripetal acceleration a_c. The magnitude of this force is:
F=ma_c
=m*v^2/R (2)
where m is the mass of the particle F and a_c are directed toward the center of curvature of the particle's path.
<em>3- The magnitude of the static frictional force between a static object and a surface is given by:</em>
f_s=u_s*n (3)
where u_s is the coefficient of kinetic friction between the object and the surface and n is the magnitude of the normal force.
Given Data
R (radius of the car's path) = 170 m
v (speed of the car) = 25 m/s
Required Data
- In part (a), we are asked to find the coefficient of static friction u_s that will prevent the car from sliding.
- In part (b), we are asked to find the speed of the car v if the coefficient of static friction is one-third u_s found in part (a).
Solution:
see the attachment pic
Since the car is not accelerating vertically, we can model it as a particle in equilibrium in the vertical direction and apply Equation (1)
∑F_y=n-mg
= mg (4)
We model the car as a particle in uniform circular motion in the horizontal direction. The force that enables the car to remain in its circular path is the force of static friction at the point of contact between road and tires. Apply Equation (2) to the horizontal direction:
∑F_x=f_s
=m*v^2/R
Substitute for f_s from Equation (3):
u_s=m*v^2/R
Substitute for n from Equation (4):
u_s*mg=m*v^2/R
u_s*g = v^2/R
Solve for u_s:
u_s= v^2/Rg (5)
Substitute numerical values:
u_s=0.375
(b)
The new coefficient of static friction between the tires and the pavement is:
u_s'=u_s/3
Substitute u_s/3 for u_s in Equation (5):
u_s/3=v^2/Rg
Solve for v:
v=14.4 m/s
