Q#13
Find the average frictional force needed to stop a car weighing 500 kg in a distance of 25 m if the initial speed is 72 km/h.Answer:
Mass m = 500 kg,
Initial speed u =72 km/h = 72000 m/3600 s = 20 m/s
Initial Kinetic Energy = ½mv² = ½ x 500 kg x (20 m/s)² = 100 kJ
Final speed v = 0, Final Kinetic Energy = 0
So work done by the average frictional force F = Change in Kinetic Energy
= 100 kJ - 0 = 100 kJ
But also work done = Force x distance = F x d (Distance d = 25 m)
= F x 25 J
So, F x 25 = 100 kJ
F = 100000J/25 m = 4000 N
(Note: This problem can also be solved by calculating acceleration/retardation. Here 0² = 20² - 2a x 25, a = 400/50 = 8 m/s²; Force = mass x acceleration = 500 x 8 = 4000 J)
Q#14
Find the average force needed to accelerate a car weighing 500 kg from rest to 72 km/h in a distance of 25 m.
Answer:
Mass m = 500 kg,
Initial speed u = 0
Initial Kinetic Energy = 0
Final speed v = 72 km/h = 72000 m/3600 s = 20 m/s,
Final Kinetic Energy = ½mv² = ½ x 500 kg x (20 m/s)² = 250 x 400 J = 100 kJ
So work done by the average force F = Change in Kinetic Energy
= 100 kJ - 0 = 100 kJ
But also work done = Force x distance = F x d (Distance d = 25 m)
= F x 25 J
So, F x 25 = 100000
F = 100000/25 = 4000 N
Q#15
A particle of mass m moves on a straight line with its velocity varying with the distance traveled according to the equation v = a√x, where a is constant, Find the total work done by all the forces during a displacement from x = 0 to x = d.
Answer:
Initial velocity u (at x = 0) = a√0 = 0
Initial Kinetic energy = ½mv² = 0
Final velocity v (at x = d) = a√d
Final Kinetic energy = ½ma²d
Work done by all the forces during the displacement d = Change in kinetic energy
∆K.E = ½ma²d – 0 = ½ma²d
Q#16
A block of mass 2.0 kg kept at rest on an inclined plane of inclination 37° is pulled up the plane by applying a constant force of 20 N parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed 40 J. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is 40 J. (c) Find the kinetic energy of the block at the instant the force ceases to act. Take g = 10 m/s².
Answer:
Let us draw a figure showing forces acting on the block
Weight = mg = 2 x 10 = 20 N, Force F = 20 N
Net force along incline = F - mgsinθ = 20 – 20 x sin37° = 7.96 N
Acceleration a = Force/mass = 7.96/2.0 = 3.98 m/s²
Distance travelled = s = ut + ½at² = 0 + 0.5 x 3.98 x1² = 1.99 m
It is the maximum work done by the force because we have not considered the frictional force i.e. when the surface is smooth otherwise it will be even less.
(b) In the above situation the upward vertical movement of block during 1 s
= 1.99 x sin37° = 1.20 m
So work done by the force of gravity (Downward) = mg(–1.20) = –24 J
(c) Since the total work done on the block = Work done by the applied force + Work done by the gravity + Work done by the Normal force
= 40 J + (–24 J) + 0 = 16 J
And Change in K.E = Final K.E.-Initial K.E. = Work done by all the forces = 16 J
So, Final K.E. = 16 J (Since Initial K.E. = 0)
Q#17
A block of mass 2.0 kg is pushed down an inclined plane of inclination 37° with a force of 20N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of 10 m/s². If the block started from rest, find the work done (a) by the applied force in the first second, (b) by the weight of the block in the first second and (c) by the frictional force acting on the block in the first second. Take g = 10 m/s².
Answer:
First let us draw the free body diagram of the block, see figure below,
The forces on the block are, Weight W = mg = 2 x 10 = 20 N (Downward)
Applied force P = 20 N (Down along the plane)
Normal force N = mgcos37° = 20 x 0.80 = 16 N (up & perpendicular to the plane)
Frictional force F = µN (Up along the plane)
Now initial speed u = 0, Acceleration a = 10 m/s², Time t = 1 s,
Final speed v = u + at = 0 + 10 x 1 = 10 m/s
And distance travelled = s = ut + ½at² = 0 + 0.5 x 10(1)2 = 5 m
(a) So work done by the applied force in 1 s = F x d =20 N x 5 m = 100 J
(b) Since the weight acts downwards, distance travelled downward = 5 Sin37° m = 3.0 m
So, work done by the gravity = 20 N x 3 m = 60 J
Or in other way, component of gravity along the plane = mgsin37° = 20 x 0.6 N = 12.0 N
Work done = 12.0 N x 5 m = 60 J
(c) Now work done by all the forces = Change in K.E.
= ½mv² - ½mu² = 0.5 x 2.0 x 10² - 0 = 100 J
Work done by the frictional force = Total W.D. by all the forces-W.D. by applied force - W.D. by the gravity - W.D. by the normal force
= 100 J – 100 J – 60 J – 0 (W.D. by the normal force is zero because movement in the direction perpendicular to plane is zero)
= –60 J
Applied force P = 20 N (Down along the plane)
Normal force N = mgcos37° = 20 x 0.80 = 16 N (up & perpendicular to the plane)
Frictional force F = µN (Up along the plane)
Now initial speed u = 0, Acceleration a = 10 m/s², Time t = 1 s,
Final speed v = u + at = 0 + 10 x 1 = 10 m/s
And distance travelled = s = ut + ½at² = 0 + 0.5 x 10(1)2 = 5 m
(a) So work done by the applied force in 1 s = F x d =20 N x 5 m = 100 J
(b) Since the weight acts downwards, distance travelled downward = 5 Sin37° m = 3.0 m
So, work done by the gravity = 20 N x 3 m = 60 J
Or in other way, component of gravity along the plane = mgsin37° = 20 x 0.6 N = 12.0 N
Work done = 12.0 N x 5 m = 60 J
(c) Now work done by all the forces = Change in K.E.
= ½mv² - ½mu² = 0.5 x 2.0 x 10² - 0 = 100 J
Work done by the frictional force = Total W.D. by all the forces-W.D. by applied force - W.D. by the gravity - W.D. by the normal force
= 100 J – 100 J – 60 J – 0 (W.D. by the normal force is zero because movement in the direction perpendicular to plane is zero)
= –60 J
Post a Comment for "Solutions to Exercises on Work and Energy HC Verma's Concepts of Physics Part 1 (13-17)"