# A 0.240 kg blob of clay is thrown at a wall with in an initial velocity of 23 m/s . If the clay comes to a stop in 91.0 ms .What is the average force experienced by the clay?

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### 1 Answer

From the equation of impulse we get:

F*t= m*v

Here, m = 0.240 kg, v= 23 m/s and t = 91.0 ms = 91*10^-3 s. Now, plugging in the values in the above relation we get:

`F * 91*10^-3 = 0.240*23 `

`rArr F = (0.240*23) /(91*10^-3)`

`rArr F=60.65934 N`

`rArr F~~61 N`

Hence, the average force experienced by the clay is **61 N**.

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