Question

Find the work W1 done on the blTwo tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.10×106 N , one at an angle 10.0 ∘ west of north, and the other at an angle 10.0 ∘ east of north, as they pull the tanker a distance 0.890 km toward the north.ock by the force of magnitude F1 = 80.0 N as the block moves from xi = -5.00 cm to xf = 6.00 cm .

Answer 1

Answer:

The work done is $3.68\times10^{9}\ J$.

Explanation:

Given that,

Force $F=2.10\times10^{6}\ N$

Angle = 10.0°

Distance = 0.890 km

Magnitude of force = 80.0 N

Initial distance = -5.00 cm

Final distance = 6.00 cm

According to figure,

The horizontal forces are same in magnitude and opposite direction.

So, neglect these two forces.

we can take only vertical components of the force.

We need to calculate the total force

$F'=F\cos10+F\cos10$

$F'=2F\cos10$.....(I)

We need to calculate the work done

Using formula of work done

$W=F\cdot s$

Put the value into the formula

$W=2\times2.10\times10^{6}\cos10\times0.890\times10^{3}$

$W=3.68\times10^{9}\ J$

Hence, The work done is $3.68\times10^{9}\ J$.