# Why am I getting wrong value here?

A man threw a ball with force 10 N to two displacement of 5 m and 10 m.

• F = ma, so let mass be 2 kg.
• a = 5 m/s2.

Work done = 10 × 5 and 10 × 10 = 50 Nm and 100 Nm.

Now, since a = 5, I assumed that after 5 m velocity = 25 [m/s].

Now, when I do K.E f - K.E initial for final work done i.e 50 Nm for now.

• Initial K.E. = 0.

So , then K.E final is not equal to work done final. 25 × 25 is not equal to 50 Nm.

Where am I wrong?

• when you wrote "Now , since a = 5, I assumed that after 5meres velocity = 25m/s" did you actually mean "*Now , since a = 5[m/s^2], I assumed that after 5[metres] the velocity will be v=25[m/s]" Jan 30 at 15:16
• I've done a major clean-up on your question. Please check that I have not mis-interpreted anything. Accurate punctuation, spacing and capitalisation is important for clarity. Jan 30 at 16:59
• "Now, since a = 5, I assumed that after 5 m velocity = 25 m/s." Why did you assume that? You need to calculate it. Hint: for how long was the ball accelerated in each case? You seem to be basing your assumption on a man with a 5 m long arm. Jan 30 at 17:09
• every line! not one word makes sense! sorry! Jan 30 at 17:54
• "It is like after every 1 sec. velocity increases by 5 m/s." It can only accelerate while it is in the thrower's hand. Once let go it won't accelerate any more (except by gravity which doesn't seem to be involved here.) So how could it accelerate for 5 s? Jan 30 at 20:31

The mistake is in your assumption.

Now, since a = 5, I assumed that after 5 m velocity = 25 m/s.

the correct would have been

Now, since a = 5, I assumed that after 5 [s] velocity = 25 m/s.

The way you can calculate it is from the following form of the equation for acceleration:

$$a = u \frac{du}{ds}$$

you can obtain that $$a\cdot 5m = \frac{u_{out}^2}{2} \rightarrow u_{out} = \sqrt{2 \cdot a\cdot 5[m]}$$.

The rest I leave to you.

• Ok.thank you my friend . Jan 31 at 4:00