Q#1
The gravitational force acting on a particle of 1g due to a similar particle is equal to 6.67 x 10-17N. Calculate the separation between the particles.Answer:
Let the separation between the particles be r. Mass of particles m=1 g =1x10-3 kg, Force between them F = 6.67 x 10-17 N,
From the gravitational equation we have
F = Gm²/r²
6.67 x 10-17 = 6.67 x 10-11 (1 x 10-3)²/r²
r² = 1
r = 1 m
Q#2
Calculate the force with which you attract the earth.
Answer:
The magnitude of force with which we attract the earth is exactly the same as the earth attracts us, that is our weight. So if my mass is m kg then this force ie my weight is mg N.
Q#3
At what distance should two charges, each equal to 1 C, be placed so that the force between them equals your weight?
Answer:
Let the distance be r.
The force between two charges
F = (1/4πε0)q1q2/r²
The term 1/4πε0 = 9 x 109 Nm²/C², q1 = q2 = 1 C, If my mass be m kg then here F= mg N, Now the equation becomes
mg= 9 x 109 x1² /r²
r² = 9 x 109/mg = 9 x 108/m
(if g = 10 m/s²)
r = 3 x 104/√m
If m = 64 kg ; r = 30000/8 m = 3.75 km.
Q#4
Two spherical bodies, each of mass 50 kg are placed at a separation of 20 cm. Equal charges are placed on the bodies and it is found that the force of Coulomb repulsion equals the gravitational attraction in magnitude. Find the magnitude of the charge placed on either body.
Answer:
r = 20 cm = 0.20 m, mass m = 50 kg each
Let q C be the equal charge on each of the spheres.
Coulomb force of repulsion = 9 x 109q²/r² N
Gravitational force of attraction = 6.67 x 10-11(50)²/r²
Equating these two we get
q²= 6.67x10-11(50)² /9 x 109 = 1853 x 10-20
q = √(1853 x 10-20) = 43.0 x 10-10 C = 4.3 x 10-9 C
Note, the separation r = 20 cm is a redundant data which cancels out in the equation.
Q#5
A monkey is sitting on a tree limb. The limb exerts a normal force of 48 N and a frictional force of 20 N. Find the magnitude of the total force exerted by the limb on the monkey.
Answer:
A diagram has been drawn for this problem on the right. The Normal force of 48 N will act perpendicular to the tree limb surface on the monkey. The force of friction 20 N acts parallel to the surface of the limb on the monkey opposing the tendency to slip. So these two forces are perpendicular to each other. The magnitude of the total force can be calculated as,
F² = 48² + 20²
F = √2704 = 52 N
Q#6
A bodybuilder exerts a force of 150 N against a bullworker and compresses it by 20 cm. Calculate the spring constant of the spring of the bullworker.
Answer:
Spring constant of a spring is the force required to compress/elongate the spring by unit length. Here 150 N force compresses the spring by 20 cm = 0.20 m.
So Spring constant of this spring = 150 N/ 0.20 m = 750 N/m.
F² = 48² + 20²
F = √2704 = 52 N
Q#6
A bodybuilder exerts a force of 150 N against a bullworker and compresses it by 20 cm. Calculate the spring constant of the spring of the bullworker.
Answer:
Spring constant of a spring is the force required to compress/elongate the spring by unit length. Here 150 N force compresses the spring by 20 cm = 0.20 m.
So Spring constant of this spring = 150 N/ 0.20 m = 750 N/m.
label: Solutions to Exercises on Forces HC Verma's Concepts of Physics Part 1,
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