Question

The position x attained by a body of mass 0.5 kg when a force acts on it vari
x = 3t² + 4t + 5 where, x is expressed in meters and t in seconds. What is the
done in joules by the force in the first 2 seconds?​

Answer 1

Answer :

• Work done on the body in 2 s, W = 60 J

Explanation :

Given :

• Position of the particle, x = 3t² + 4t + 5
• Mass of the body, m = 0.5 kg

To find :

• Work done on the body at an instànt of 2 s, W = ?

Knowledge required :

• Differentiation of the position of a particle gives the velocity of that particle.

Formula for velocity of a particle, v = d(x)/dt.

• Differentiation of the velocity of a particle gives the acceleration of that particle.

Formula for acceleration of a particle, a = d(v)/dt.

• Power rule of differentiation : d(x^n)/dx = n × x^(n - 1)
• Constant rule of differentiation : (c)/dx = 0 [Where, c = constant]

• Formula for force exerted by a body, F = ma.

[Where: F , m and a are force exerted by the body, mass of the body and acceleration of the body, respectively]

• Formula for Work done by the body, W = F s

[Where: W, F and s are work done by the body, force exerted and displacement of the body]

Solution :

To find the displacement of the particle in the first two seconds :

⠀By substituting the value of t in the position of the body, we get :

⠀⠀=> x₍₀ ₜₒ ₂₎ = x₂ - x₁

⠀⠀=> x₂ - x₁ = [3(2)² + 4(2) + 5] - [3(0)² + 4(0) - 5]

⠀⠀=> x₂ - x₁ = [3(2)² + 4(2) + 5] - [3(0)² + 4(0) - 5]

⠀⠀=> x₂ - x₁ = (12 + 8 + 5) - 5

⠀⠀=> x₂ - x₁ = 25 - 5

⠀⠀⠀⠀∴ x₍ₜ ₌ ₂ ₛ₎ = 20 m

Hence the position or displacement of the body in 2 s is 20 m.

To find the velocity of the particle :

⠀⠀By using the formula for velocity of a particle and differentiating it with respect to time, we get :

⠀⠀=> v = d(x)/dt

⠀⠀=> v = d(x)/dt = d(3t² + 4t + 5)/dt

⠀⠀=> v = d(x)/dt = d(3t²)/dt + d(4t) + d(5)/dt

⠀⠀=> v = d(x)/dt = [2 × 3t⁽² ⁻ ¹⁾] + [1 × 4t⁽¹ ⁻ ¹⁾] + 0

⠀⠀=> v = d(x)/dt = 2 × 3t¹ + [1 × 4t⁰]

⠀⠀=> v = d(x)/dt = 6t + 4

⠀⠀⠀⠀∴ v = (6t + 4) m/s

Hence the velocity of the body is (6t + 4) m/s.

To find the acceleration of the body :

⠀⠀By using the formula for acceleration of a particle and differentiating it with respect to time, we get :

⠀⠀=> a = d(v)/dt

⠀⠀=> a = d(v)/dt = d(6t + 4)/dt

⠀⠀=> a = d(v)/dt = d(6t)/dt + d(4)/dt

⠀⠀=> a = d(v)/dt = [1 × 6t⁽¹ ⁻ ¹⁾] + 0

⠀⠀=> a = d(v)/dt = [1 × 6t⁰]

⠀⠀=> a = d(v)/dt = 6

⠀⠀⠀⠀∴ a = 6 m/s²

Hence the acceleration of the body is 6 m/s².

Hence the acceleration of the body at the instànt of 2 s is 6 m/s².

To find the force acting on the body :

⠀By using the formula for force exerted by a body and substituting the values in it, we get :

⠀⠀=> F = ma

⠀⠀=> F = 0.5 × 6

⠀⠀=> F = 3

⠀⠀⠀∴ F = 3 N

Hence the force acting on the body is 3 N.

To find the work done by the body :

⠀By using the formula for work done on the body and substituting the values in it, we get :

⠀⠀=> W = F s

⠀⠀=> W = 3 × 20

⠀⠀=> W = 60

⠀⠀⠀∴ W = 60 J

Hence the work on the body is 60 J.

Answer 2

$\huge{\underline{\mathtt{\red{H}\pink{E}\green{L}\blue{L}\orange{O}}}}$

To find the displacement of the particle in the first two seconds :

⠀By substituting the value of t in the position of the body, we get :

⠀⠀=> x₍₀ ₜₒ ₂₎ = x₂ - x₁

⠀⠀=> x₂ - x₁ = [3(2)² + 4(2) + 5] - [3(0)² + 4(0) - 5]

⠀⠀=> x₂ - x₁ = [3(2)² + 4(2) + 5] - [3(0)² + 4(0) - 5]

⠀⠀=> x₂ - x₁ = (12 + 8 + 5) - 5

⠀⠀=> x₂ - x₁ = 25 - 5

⠀⠀⠀⠀∴ x₍ₜ ₌ ₂ ₛ₎ = 20 m

Hence the position or displacement of the body in 2 s is 20 m.

$\pink{To find the velocity of the particle :}$

⠀⠀By using the formula for velocity of a particle and differentiating it with respect to time, we get :

⠀⠀=> v = d(x)/dt

⠀⠀=> v = d(x)/dt = d(3t² + 4t + 5)/dt

⠀⠀=> v = d(x)/dt = d(3t²)/dt + d(4t) + d(5)/dt

⠀⠀=> v = d(x)/dt = [2 × 3t⁽² ⁻ ¹⁾] + [1 × 4t⁽¹ ⁻ ¹⁾] + 0

⠀⠀=> v = d(x)/dt = 2 × 3t¹ + [1 × 4t⁰]

⠀⠀=> v = d(x)/dt = 6t + 4

⠀⠀⠀⠀∴ v = (6t + 4) m/s

Hence the velocity of the body is (6t + 4) m/s.

$\blue{To find the acceleration of the body :}$

⠀⠀By using the formula for acceleration of a particle and differentiating it with respect to time, we get :

⠀⠀=> a = d(v)/dt

⠀⠀=> a = d(v)/dt = d(6t + 4)/dt

⠀⠀=> a = d(v)/dt = d(6t)/dt + d(4)/dt

⠀⠀=> a = d(v)/dt = [1 × 6t⁽¹ ⁻ ¹⁾] + 0

⠀⠀=> a = d(v)/dt = [1 × 6t⁰]

⠀⠀=> a = d(v)/dt = 6

⠀⠀⠀⠀∴ a = 6 m/s²

Hence the acceleration of the body is 6 m/s².

Hence the acceleration of the body at the instànt of 2 s is 6 m/s².

$\green{To find the force acting on the body :}$

⠀By using the formula for force exerted by a body and substituting the values in it, we get :

⠀⠀=> F = ma

⠀⠀=> F = 0.5 × 6

⠀⠀=> F = 3

⠀⠀⠀∴ F = 3 N

Hence the force acting on the body is 3 N.

$\green{To find the work done by the body :}$

⠀By using the formula for work done on the body and substituting the values in it, we get :

⠀⠀=> W = F s

⠀⠀=> W = 3 × 20

⠀⠀=> W = 60

⠀⠀⠀∴ W = 60 J

Hence the work on the body is 60 J.

✰✫✰✫✰✫✰✫✰✫✰✫

▸▸▸♪