Exercise - 1

Given:
Position vector, r = (10t)i + (20t²)j

Step 1: Find the velocity.
v = dr/dt
v = 10i + 40t j

Step 2: Find the acceleration.
a = dv/dt
a = 40j m/s²

The acceleration has no x-component and is entirely in the +y direction.

Step 3: Use Newton's second law.
F = ma

Since acceleration is in the +y direction, the net force is also in the +y direction.

Therefore, the direction of the net force is the +y-axis.

Given:
F₁ = 8 N
F₂ = 6 N
Mass of the ball, m = 5 kg

Step 1: Find the resultant force.
Since F₁ and F₂ are perpendicular:
F_net = √(F₁² + F₂²)
F_net = √(8² + 6²)
F_net = √(64 + 36)
F_net = √100
F_net = 10 N

Step 2: Use Newton's second law.
F_net = ma
a = F_net / m
a = 10 / 5
a = 2 m/s²

Therefore, the acceleration of the ball is 2 m/s².

Given:
F₁ = 10 N
F₂ = 8 N
F₃ = 6 N
F₂ and F₃ are perpendicular to each other.
The resultant of F₂ and F₃ is opposite to F₁.

Step 1: Find the resultant of F₂ and F₃.
Since F₂ and F₃ are perpendicular:
R = √(F₂² + F₃²)
R = √(8² + 6²)
R = √(64 + 36)
R = √100
R = 10 N

Step 2: Compare the resultant with F₁.
The resultant of F₂ and F₃ is 10 N and is opposite to F₁, which is also 10 N.

Therefore, the two resultants cancel each other:
F_net = F₁ + R
F_net = 10 - 10
F_net = 0 N

Therefore, the net force on the particle is 0 N.

Given:
F₁ = 10 N
F₂ = 8 N
F₃ = 6 N
Mass of the ball, m = 5 kg

Since F₁ is removed, only F₂ and F₃ act on the ball.

Step 1: Find the net force.
F₂ and F₃ are perpendicular, so their resultant is:
F_net = √(F₂² + F₃²)
F_net = √(8² + 6²)
F_net = √(64 + 36)
F_net = √100
F_net = 10 N

Step 2: Find the acceleration using Newton's second law.
F_net = ma
a = F_net / m
a = 10 / 5
a = 2 m/s²

Therefore:
Net force = 10 N
Acceleration = 2 m/s²

The net force and acceleration act in the direction of the resultant of F₂ and F₃.

Given:
Mass of the ball, m = 1 kg
Velocity, v = 10√x m/s

Since velocity is given as a function of position, use:
a = v(dv/dx)

Step 1: Differentiate velocity with respect to x.
dv/dx = d(10√x)/dx
dv/dx = 5/√x

Step 2: Find the acceleration.
a = v(dv/dx)
a = (10√x)(5/√x)
a = 50 m/s²

Step 3: Use Newton's second law.
F_net = ma
F_net = 1 × 50
F_net = 50 N

Therefore, the net force acting on the ball is 50 N in the positive x-direction.

Given:
Mass of the ball, m = 1 kg
Velocity, v = (2t)î + (3t²)ĵ m/s
Time, t = 1 s

Step 1: Find the acceleration by differentiating velocity with respect to time.
a = dv/dt
a = 2î + 6t ĵ

Step 2: Substitute t = 1 s.
a = 2î + 6ĵ m/s²

Step 3: Use Newton's second law.
F = ma
F = 1(2î + 6ĵ)
F = 2î + 6ĵ N

Therefore, the force acting on the ball at t = 1 s is F = 2î + 6ĵ N.

Given:
Mass of the ball, m = 20 kg
Initial velocity, u = 0 m/s
Final velocity, v = 5 m/s
Time, t = 20 s

Step 1: Find the acceleration using:
v = u + at
5 = 0 + a(20)
a = 5/20
a = 0.25 m/s²

Step 2: Use Newton's second law.
F = ma
F = 20 × 0.25
F = 5 N

Therefore, the force acting on the ball is 5 N.

Given:
Mass of the ball, m = 20 kg
Time for which force acts, t₁ = 20 s
Initial velocity, u = 0 m/s
Distance travelled after the force is removed, s = 50 m
Time after the force is removed, t₂ = 10 s

After the force is removed, the ball moves with constant velocity.

Step 1: Find the velocity of the ball when the force is removed.
v = s/t
v = 50/10
v = 5 m/s

Therefore, the ball reaches a velocity of 5 m/s during the 20 s in which the force acts.

Step 2: Find the acceleration while the force acts.
v = u + at
5 = 0 + a(20)
a = 5/20
a = 0.25 m/s²

Step 3: Use Newton's second law.
F = ma
F = 20 × 0.25
F = 5 N

Therefore, the force acting on the ball is 5 N.

Given:
Mass of the ball, m = 70 kg
Horizontal force, F = 200 N
Acceleration due to gravity, g = 10 m/s²

The applied force acts only in the horizontal (+x) direction, so it has no vertical component.

Step 1: Find the weight of the ball.
W = mg
W = 70 × 10
W = 700 N

Step 2: Find the normal force.
There is no vertical acceleration, so the normal force balances the weight.
N = W
N = 700 N

Therefore, the normal force acting on the ball is 700 N.