A body of mass 2\text{ kg} is projected vertically upward from ground level with an initia… — JAMB UTME

A body of mass $2\text{ kg}$ is projected vertically upward from ground level with an initial speed of $30\text{ m s}^{-1}$ . During its entire flight, it experiences a constant resistive force due to air resistance of $5\text{ N}$ . Taking $g = 10\text{ m s}^{-2}$ , calculate the kinetic energy of the body in Joules when it returns to the ground level.

Answer: 540 J

Answer

The kinetic energy of the body when it returns to ground level is

540\text{ J}

At launch, the body possesses an initial kinetic energy of

E_i = \frac{1}{2}(2)(30)^2 = 900\text{ J}

. During ascent, both gravity (

20\text{ N}

) and air resistance (

5\text{ N}

) retard the motion, giving a net downward force of

25\text{ N}

and a deceleration of

12.5\text{ m s}^{-2}

. The maximum height reached is

h = \frac{30^2}{2(12.5)} = 36\text{ m}

. Since the body travels up and down, total distance covered is

72\text{ m}

. Non-conservative work done against air resistance is

W = 5\text{ N} \times 72\text{ m} = 360\text{ J}

. Therefore, the remaining kinetic energy upon returning to the ground is

900\text{ J} - 360\text{ J} = 540\text{ J}

Step-by-Step Solution

Calculate initial kinetic energy of launch

E_i = 900\text{ J}

Kinetic energy is given by

\frac{1}{2}m u^2 = \frac{1}{2}(2\text{ kg})(30\text{ m s}^{-1})^2 = 900\text{ J}

Determine maximum height reached during ascent

h = 36\text{ m}

Net upward retarding force

F = mg + F_{\text{air}} = 20 + 5 = 25\text{ N}

, yielding a deceleration

a = 12.5\text{ m s}^{-2}

. Using

0 = u^2 - 2ah

h = \frac{900}{25} = 36\text{ m}

Calculate energy lost to air resistance over total trajectory

W_{\text{air}} = 360\text{ J}

Air resistance acts continuously over both ascent and descent (total distance

2h = 72\text{ m}

). Work dissipated

= F_{\text{air}} \times 2h = 5 \times 72 = 360\text{ J}

Subtract non-conservative work loss from initial mechanical energy

E_f = 540\text{ J}

By mechanical energy balance, final kinetic energy

E_f = E_i - W_{\text{air}} = 900 - 360 = 540\text{ J}

Key Concept

Work-Energy Theorem and Mechanical Energy Dissipation by Non-Conservative Forces

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic