Question

Bob, who has a mass of 79 kg , can throw a 510 g rock with a speed of 34 m/s . The distance through which his hand moves as he accelerates the rock forward from rest until he releases it is 1.0 m. Part A What constant force must Bob exert on the rock to throw it with this speed?

Answer 1

F=294.78 N

Step-by-step explanation:

Given that

M= 79 kg

m= 510 g= 0.51 kg

Initial speed u = 0 m/s

Final speed ,v= 34 m/s

d= 1 m

We know that

v²=u²+ 2 a d

By putting the values

34²= 0 ² + 2 x a x 1

a= 578 m/s²

From second law of Newton's

F= m a

By putting the values

F= 0.51 x 578 N

F=294.78 N

Answer 2

Force will be 2947.8 N

Step-by-step explanation:

We have given that mass of the rock m = 510 gram = 0.510 kg

Final velocity v = 34 m/sec

Distance s = 1 m

Initial velocity u = 0 m/sec

From third equation of motion we know that
`v^2=u^2+2as`

`34^2=0^2+2* a* 1`

`a=578m/sec^2`

From newton's law of motion we know that F = ma

So force
`F=0.510* 578=2947.8N`