Question

a force of 5 Newton making an angle with the horizontal acting on an object is place is it by 0.4 along the horizontal direction
if the object gains kinetic energy of an joul horizontal component of the force is​

Answer 1

Answer:

2.5Newtons

Explanation:

Let the force make θ with horizontal. Then Fcosθ is its component along horizontal.

Now,

F=ma

Fcosθ=ma

a=Fcosθ/m

Let Fcosθ be x.

a=x/m                  ---1

Using Newton's Second eq:

$s=ut+\frac{at^2}{2}\\0.4=0+\frac{xt^2}{2m}$         ---2

Now given that Kinetic energy is 1 J.

mv²=2*1

v=√(2/m)

Now use first eq of motion:

$v=u+at\\\sqrt{\frac{2}{m}} = 0+\frac{x}{m} t\\\sqrt{\frac{2}{m}}=\frac{x}{m}t\\t=\frac{\sqrt{2m}}{x}$

Use this value in 2:

$0.4=xt^2/2m\\\frac{4}{10}=(x(\frac{\sqrt{2m}}{x}))^2/2m\\\frac{4}{10}=(x(\frac{2m}{x^2}))/2m\\\frac{4}{10}=\frac{1}{x}\\x=10/4\\x=2.5 N$

Hence, Horizontal component is 2.5N