Question on Co-efficient of Friction

Friction is a force which always acts to oppose the motion (any type of motion) of an object over a surface and is an example of a contact force.

Topic : Coefficient of friction

Coefficient of friction is given by V²/(radius * gravity)

Question : A car of mass = 1400 kg traveling at 65.0 km/h enters a banked turn covered with ice. The road is banked at an angle , and there is no friction between the road and the car’s tires. Now, suppose that the curve is level and that the ice has melted, so that there is a coefficient of static friction between the road and the car’s tires. What is, the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car’s speed is still 65.0 and that the radius is 91.3m?

Solution :

Mass = 1400 kg

Velocity = 65.0 km/h

= 18.07 m/sec

Radius = 91.3m

V² = co-efficient of friction * radius * g

Coefficient of friction = V²/ (radius * g)

= 18.07²/ (91.3*9.8)

= 0.365

The minimum value of the coefficient of static friction =0.365

 

For more number of questions for practice is found in physics help.

 

Question On Acceleration

Topic: Acceleration

Question: A lift  of mass 400kg is hung by a wire. Calculate the tension in the wire when the lift is

a) at rest

b) moving upward with a constant  velocity of 2 m/s

c) moving upward with an acceleration of 2 m/s^2

d) moving downward with an acceleration of 2 m/s^2

Solution:

a) When lift is at rest ,then  T = Mg = 400 * 9.8 = 3920 N
b)When lift is moving upward with constant velocity , acceleration is zero,

so T – Mg= M*0  or ,  T = Mg =  400 * 9.8 = 3920 N

c) When lift  moves up with acceleration a, then

T – Mg = Ma
T= Mg+Ma = M( g+a)
T= 400 (9.8+2) = 400*11.8 = 4720 N

d) When lift moves downward with acceleration a, then
Mg- T = Ma
T  = Mg –Ma= M(g-a)
T  = 400 ( 9.8-2)  = 400 * 7.8 = 3120 N

 

Question on Finding Specific Heat of a Constant Volume

Topic : Specific Heat

 

Question : The velocity of sound in an ideal gas is,

v=(γP/ρ)^1/2 . Find the value of specific heat at constant

volume.

Solution:

The ratio of the specific heat is γ

c_p/c_v = γ =1.5

c_p = 1.5 c_{v ————————   (1)}

Also c_p-c_v=R. Substituting the value of c_p from equation 1,

1.5c_v - c_v = R => .5c_v=R => c_v = R/.5 = 2R

 

 

 

 Hope derivation to the value of Specific heat was satisfied.