QJEEM23Ph9d08: JEE Main 2023 Physics - Class 11 JEEM23Ph9d08 Solution: A block of mass 5   kg starting from rest pulled up on a…

To find the power delivered by the pulling force at t = 10 s, we first need to find the work done by the force. The work done is given by the product of force and displacement, and the power is the rate of work done.

Calculate the velocity (v) at t = 10 s:

Since the block starts from rest and is pulled up with an effective acceleration of 1 m/s², we can use the equation of motion to find the velocity (v) at t = 10 s:

$$v=u+at$$

Here, u = 0 (initial velocity) and a = 1 m/s² (acceleration). Plugging in the values:

$$v_{10}=0+1(10)=10\mathrm{m}/\mathrm{s}$$

1. Calculate the net force acting on the block ( $F_{net}$ ):
   
   The net force acting on the block along the incline plane is the difference between the pulling force (F) and the gravitational force component acting parallel to the incline (mgsinθ):

$$F_{\text{net}}=F-mgsin\theta$$

Since F_net = ma, we can write:

$$F=ma+mgsin\theta$$

Plugging in the values (m = 5 kg, a = 1 m/s², g = 10 m/s², and θ = 30°):

$$F=5(1)+5(10)(\sin 30°)=5+25=30\mathrm{N}$$

Calculate the power (P) at t = 10 s:

The power (P) can be calculated as the product of force (F) and velocity (v):

$$P_{10}=Fv=30(10)=300\mathrm{W}$$

So, the power delivered by the pulling force at t = 10 s from the start is 300 W.