Question

Find the speed of the block when it covers a horizontal distance { . It is given that the block never loses contact with
the smooth horizontal surface and the force always acts at an angle with the horizontal,
Fo
U=0
M
smooth​

Answer 1

Answer:

1. answer is the image
2. find the speed

Answer 2

Answer:

The speed of the block is √(2sFcosθ/m) .

Explanation:

Given that :

• Force, F acts at an angle θ with horizontal
• Surface is smooth and horizontal

To find :

• Speed of the block

Solution :

• Let, the mass of the block be m.
• Considering, that on applying the force, F, the block moves with an uniform acceleration, a and covers distance, s.
• Since, F acts at an angle θ with horizontal, the component of force parallel to horizontal is F cosθ .
• Now, from 2nd law of motion ;

$F cosθ = ma \: \: \: \: - - - (1)$

• Using 3rd equation of motion, we get;

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

• Putting this value in equation (1), we get;

$F cosθ = \frac{m {v}^{2} }{2s} \\ {v}^{2} = {\frac{2sF cosθ}{m} } \\ v= \sqrt{{\frac{2sF cosθ}{m} }}$

• Hence, speed of the block is √(2sFcosθ/m) .