Bob, who has a mass of 60 , can throw a 800 rock with a speed of 26 . The distance through which his hand moves as he accelerates the rock forward from rest until he releases it is 1.0 .

a) What constant force must Bob exert on the rock to throw it with this speed?

b)If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?

You need to provide dimensions with your numbers.

b. F = m*a = 0.8 * 338 = 270.4 N.

To answer these questions, we can use Newton's second law of motion and principles of conservation of momentum.

a) To calculate the constant force Bob must exert on the rock, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

First, let's calculate the acceleration of the rock. We know that the distance through which Bob's hand moves is 1.0 m and the final velocity of the rock is 26 m/s. We can use the following kinematic equation:

v^2 = u^2 + 2as

where:
v = final velocity = 26 m/s
u = initial velocity = 0 m/s (since the rock was at rest)
a = acceleration (unknown)
s = distance = 1.0 m

Rearranging the equation, we get:

a = (v^2 - u^2) / (2s)

Substituting the given values:

a = (26^2 - 0^2) / (2 * 1.0)
= 676 / 2
= 338 m/s^2

Now, we can calculate the force using Newton's second law:

F = ma

Substituting the mass of the rock:

F = 800 * 338
= 270,400 Newtons

Therefore, Bob must exert a constant force of 270,400 Newtons on the rock to throw it with a speed of 26 m/s.

b) To determine Bob's recoil speed after releasing the rock on frictionless ice, we can use the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant.

The momentum of the rock before it is released is given by:

initial momentum = mass * initial velocity = 800 kg * 0 m/s = 0 kg*m/s

Since there are no external forces acting on the system after the rock is released, the total momentum must remain constant. Therefore, the momentum after the release of the rock will be opposite in direction but equal in magnitude to the initial momentum of the rock.

Since Bob's mass is 60 kg, his recoil speed can be calculated using the equation:

momentum = mass * velocity

Rearranging the equation to solve for velocity:

velocity = momentum / mass

Substituting the values:

velocity = 0 kg*m/s / 60 kg
= 0 m/s

Therefore, Bob's recoil speed will be 0 m/s on the frictionless ice after releasing the rock.

m1=60 kg, m2=0.8 kg, v=26m/s, s=1 m

(a). a=v²/2s =26²/2•1=338 m/s²
F=m1•a=60•338=20280 N (inhumanly large force... I believe that velocity has to be 2.6 m/s .... check your given data)
(b)The law of conservation of linear momentum
0= m1•v1-m2•v2
v1= m2•v2/m1= 0.8•26/60=0.347 m/s