Question

A 5kg is lifted vertically at a constant velocity of 4m/s through a height of 12m.
a) How great a force is required?
b) How much work is done? What becomes 5hr work?

Answer 1

Explanation:

v = 4m/sec

u = 0m/sec

s = 12m

${ v}^{2} = {u}^{2} + 2as \\ {4 }^{2} = {0}^{2} + 2a12 \\ 16 = 0 + 24a \\ 16 = 24 a \\ a = 24 \div 16 \\ a = 1.5m \: per \: {sec }^{2} \\ f = mass\times accelertion \\ 5 \times 1.5 \\ = 7.5n \\ work \: done = force \times distnce \\ 7.5 \times 12 \\ = 90joule$

Answer 2

Answer:

Explanation:

Given :

• Mass (m) = 5 kg
• Initial velocity (u) = 0 m/s
• Final velocity (v) = 4 m/s
• Displacement (s) = 12m

To Find :

• a) How great a force is required?
• b) How much work is done?

Formula to be used :

• v^2 = u^2 + 2as
• F = ma
• W = F × s

Solution :

★How great a force is required?

v^2 = u^2 + 2as

⇒ 4^2 = 0^2 + 2 × a × 12

⇒ 16 = 24a

⇒ a = 16/24

⇒ a = 0.67 m/s^2

Now, apply force formula,

F = ma

⇒ F = 5 × 0.67

⇒ F = 3.35 N★

★How much work is done?

W = F × s

⇒ W = 3.35 × 12

⇒ W = 40.2 J★