Question

A force acts on a 3.0 g particle in such a way
that the position of the particle as a function of
time is given by x = 3t -4t2 + tº, where x is in
metre and tis in second. The work done during
the first 4 sis :​

Answer 1

Question :

A force acts on a 3.0 g particle in such a way that the position of the particle as a function of time is given by, x = 3t - 4t2 + t³, where x is in metre and t is in second. The work done during the first four seconds.

Answer :

• The work done by the particle during at t = 4 s is 0.576 J.

Explanation :

Given :

• Position of the particle, x = 3t - 4t² + t³.
• Instant of time, t = 4 s
• Mass of the particle, m = 3 g or 0.003 kg

To find :

• Work done by the particle at t = 4s , W = ?

Knowledge required :

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀v = d(x)/dt⠀

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

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀a = d(v)/dt⠀

• Formula for work done by a particle :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀W = F s⠀

[Where : W = Work done by the particle, F = Force applied by a particle, s = Displacement of the particle]

• Formula for force of a particle :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀F = ma⠀

[Where : F = Force applied by a particle, m = mass of the particle, a = Acceleration of the particle]

• Exponent rule of differentiation :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀d(x^n)/dx = nx^(n - 1)⠀

• Differentiation of a constant term is 0.

Solution :

To find the displacement of the particle, at = 4 s :

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

⠀⠀=> s = 3t - 4t² + t³

⠀⠀=> s₍ₜ ₌ ₄ ₛ₎ = 3(4) - 4(4)² + (4)³

⠀⠀=> s₍ₜ ₌ ₄ ₛ₎ = 12 - 64 + 64

⠀⠀=> s₍ₜ ₌ ₄ ₛ₎ = 12

⠀⠀⠀⠀⠀∴ s₍ₜ ₌ ₄ ₛ₎ = 12 m

Hence the displacement of the particle, at = 4 s is 12 m.

To find the velocity of the particle :

⠀By using the formula for velocity of a particle and substituting the values in it, we get :

⠀⠀=> v = d(x)/dt

⠀⠀=> v = d(x)/dt = d(3t - 4t² + t³)/dt

⠀⠀=> v = d(x)/dt = d(3t)/dt - d(4t²)/dt + d(t³)/dt

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

⠀⠀=> v = d(x)/dt = [1 × 3t⁰] - [2 × 4t¹] + [3 × t²]

⠀⠀=> v = d(x)/dt = 3 - 8t + 3t²

⠀⠀⠀⠀⠀∴ v = 3 - 8t + 3t² m/s

Hence the velocity of the particle is 3t - 8t + 3t² m/s.

To find the acceleration of the particle :

⠀By using the formula for accelaration of a particle and substituting the values in it, we get :

⠀⠀=> a = d(v)/dt

⠀⠀=> a = d(x)/dt = d(3 - 8t + 3t²)/dt

⠀⠀=> a = d(x)/dt = d(3)/dt - d(8t)/dt + d(3t²)/dt

⠀⠀=> a = d(x)/dt = 0 - [1 × 8t⁽¹ ⁻ ¹⁾] + [2 × 3t⁽² ⁻ ¹⁾]

⠀⠀=> a = d(x)/dt = 0 - [1 × 8t⁰] + [2 × 3t¹]

⠀⠀=> a = d(x)/dt = - 8 + 6t

⠀⠀⠀⠀⠀∴ a = 6t - 8 m/s²

Hence the acceleration of the particle is 6t - 8 m/s².

To find the acceleration of the particle, at = 4 s :

⠀By substituting the value of t in the acceleration of the particle, we get :

⠀⠀=> a = 6t - 8

⠀⠀=> a₍ₜ ₌ ₄ ₛ₎ = 6(4) - 8

⠀⠀=> a₍ₜ ₌ ₄ ₛ₎ = 24 - 8

⠀⠀=> a₍ₜ ₌ ₄ ₛ₎ = 16

⠀⠀⠀⠀⠀∴ a₍ₜ ₌ ₄ ₛ₎ = 16 m/s²

Hence the acceleration of the particle, at = 4 s is 16 m/s².

To find the force exerted by the particle :

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

⠀⠀=> F = ma

⠀⠀=> F = 0.003 × 16

⠀⠀=> F = 0.048

⠀⠀⠀⠀⠀∴ F = 0.048 N

Hence the force applied by the particle is 0.048 N.

To find the work done by the particle :

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

⠀⠀=> W = F s

⠀⠀=> W = 0.048 × 12

⠀⠀=> W = 0.576

⠀⠀⠀⠀⠀∴ W = 0.576 J

Hence the work done by the particle is 0.576 N.