Question

A 50 – kg sack of rice is lifted up a height of 1.50 m. a 45.0 kg sack of corn is lifted up a 0.76 m – high table. Which sack would require more work to lift ? with computation plss​

Answer 1

$\red{\mathtt{Work \: Done = F × s}} \\ \\ \mathtt {= mgh}$

Given :

⇒For sack 1

m1 = 50 kg

h1 = 1.5m

⇒For Sack 2

m2 = 45 kg

h2 = 0.76m

Solution :-

W1 = m1gh1

W1 = 50 × 10 × 1.5

W1 = 750 Nm

W2 = m1gh2

W2 = 45 × 10 × 0.76

W2 = 342 Nm

• Therefore

Sack 1 will require more work as compared to Sack 1

Answer 2

Answer:

$\huge\mathfrak \red {solution}$

Given :

⇒For sack 1

m1 = 50 kg

h1 = 1.5m

⇒For Sack 2

m2 = 45 kg

h2 = 0.76m

Solution :-

W1 = m1gh1

W1 = 50 × 10 × 1.5

W1 = 750 Nm

W2 = m1gh2

W2 = 45 × 10 × 0.76

W2 = 342 Nm

Therefore,

Sack 1 will require more work as compared to Sack 1