Can someone please help me with this
1 answer:
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The motor does 20 J of work on the block, it means that the total mechanical energy of the block has increased by 20 J. But the increase in total mechanical energy is equal to the sum of the increases in potential energy and kinetic energy:
So, if the gravitational potential energy has increased by 15 J, the kinetic energy has increased by
So, if the gravitational potential energy has increased by 15 J, the kinetic energy has increased by
The charge on the droplet must be
Explanation:
In order for the oil droplet to be in equilibrium, the force of gravity acting on it (the weight, acting downward) must be equal to the electrical force acting upward.
The gravitational force is:
which is equivalent to the weight of the droplet.
The electric force is given by:
where
q is the charge on the oil droplet
is the magnitude of the electric field
Since the two forces must be equal,
And solving for q,
Learn more about gravitational force and electric force:
#LearnwithBrainly
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
