Question

A BLOCK OF MASS 10KG IS PULLED 20M UP A SMOOTH PLANE AT 45°TO THE HORIZONTAL . THE BLOCK IS INITIALLY AT REST AND REACHES A VELOCITY OF 2•0M/S AT THE TOP OF THE PLANE . CALCULATE THE MAGNITUDE OF THE FORCE REQUIRED ASSUMING IT ACTS PARALLEL TO THE PLANE.

PLEASE TELL THE ANSWER PLEASE ​

Answer 1

Force applied = 50.7 Newtons.

Explanation

v=2m/s

u=0

s=20m

Use second equation of motion:

$\frac{v^{2}-u^2}{2s}=acceleration$

$\frac{4}{40} = 0.1 ms^{-2}$

This acceleration is required along the incline to get v=2m/s at top of the incline.

Taking components:

Along the incline, -g sin(45°)= 10/√2 ------(1)  (Negative sign because it opposes motion)

and a sec 45 (=a√2) due to force applied. -----(2)

Net sum of accelerations 1 and 2 must be 0.1 m/s²

$\frac{-10}{\sqrt{2} } + a\sqrt{2} = 0.1$

a= 5.07 m/s²

Force applied in horizontal (parallel to the plane)

=  ma

= 10kg*5.07

=50.7 Newtons.