1500 kg automobile is traveling up to 20 degree incline at a speed of 6 m/s. if the driver wishes to stop his car in a distance of 5m, determine the frictional force at pavement which must be supplied by rear wheels
A body is said to be perfectly elastic, if it returns back___________
The kinetic energy of the car will be dissipated as heat due to the frictional forces.
KE=(1/2)mV^2=1500*36/2=27000 J
The work done by frictional forces=F*d
Now, F*d=KE
F=KE/d=27000/5=5400 N
m=1500 kg; α =20º; s=5 m
KE=W(fr)+PE.
KE=m•v²/2;
W(fr) =F(fr) •s
PE=m•g•h=m•g•s•sinα.
m•v²/2 = F(fr) •s+ m•g•s•sinα;
F(fr)= m{ v²/2 - g•s•sinα}/s =
= 1500{36/2 -9.8•5•0.342}/5 =372.6 N
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The body is perfectly elastic if after being deformed under the action of an external applied force the body completely returns to its initial condition on the removal of the force.
true or false
a)a rigid body is an idealization of the behaviour of a body.
b)A GEO stationary satelite follows a circular orbit at a unique and has a time period of 24th.
c)constancy of areal velocity for a central force motion of a particle does not imply that the angular momentum is conserved
d)every planet moves in an orbit which is an ellipse with the sun at one focus
e)for perfectly elastic central impact the two bodies must move stuck together onlu if the impact is direct
fill in the blanks----
a)the power developed by a force is ______ and that of couple is _______.
b)the velocity of cylinder which is released from rest on 0 degree incline after it has rolled through a distance l is given.
c)the conditions for non slipping is_______.
c)the direction of coriolis acceleration is obtaind by ______.
d)the action of a spring on a movable body to which it is attached is a common example of ________.
true or false
a)a rigid body is an idealization of the behaviour of a body.
b)A GEO stationary satelite follows a circular orbit at a unique and has a time period of 24th.
c)constancy of areal velocity for a central force motion of a particle does not imply that the angular momentum is conserved
d)every planet moves in an orbit which is an ellipse with the sun at one focus
e)for perfectly elastic central impact the two bodies must move stuck together onlu if the impact is direct
To determine the frictional force required to stop the car in a distance of 5m, we need to consider the forces acting on the car.
1. Weight (W): The weight of the car is the force due to gravity and acts vertically downwards. It can be calculated as the mass of the car (1500 kg) multiplied by the acceleration due to gravity (9.8 m/s²). Therefore, W = 1500 kg × 9.8 m/s².
2. Normal force (N): The normal force is the force exerted by the surface (pavement) on the car perpendicular to the incline. It can be calculated by resolving the weight of the car along and normal to the incline. Since the incline is 20 degrees, the normal force can be calculated as N = W × cos(20°).
3. Frictional force (Ff): The frictional force opposes the motion of the car. In this case, it acts in the direction opposite to the motion of the car. The frictional force required to stop the car can be calculated using the formula Ff = mass × acceleration.
Now, we need to calculate the acceleration:
To calculate the acceleration, we need to find the net force acting on the car in the direction opposite to the motion. The net force can be calculated as the difference between the frictional force and the component of the weight parallel to the incline.
The component of the weight parallel to the incline can be calculated as W_parallel = W × sin(20°).
Now, the net force can be calculated as F_net = Ff - W_parallel.
Since the car is brought to a stop, the final velocity (v) will be 0 m/s. The initial velocity (u) is given as 6 m/s, and the distance traveled (s) is given as 5 m.
We can calculate the acceleration using the equation v² = u² + 2as. Substituting the values, we have:
0 = (6 m/s)² + 2a × 5 m.
Solving for acceleration, we get a = -18 m/s² (negative sign indicates deceleration).
Now that we have the acceleration, we can calculate the net force:
F_net = Ff - W_parallel = mass × acceleration.
Substituting the values, we have:
Ff - W_parallel = 1500 kg × (-18 m/s²).
Finally, we can calculate the frictional force:
Ff = W_parallel - 1500 kg × 18 m/s².