a rock is thrown upward at a speed of 10 m/s, the force needed to bring it to a stop in 10s is:?

Answer 1

a = 1 m/s2
m= x kg
F = x kg m / s2

Answer 2

No way to know unless the mass is given.

In terms of acceleration:

vf = vi + at

where vf is final velocity, vi is initial velocity, a is acceleration, and t is time.

So:
0 = 10 + a(10)

Net acceleration has to be -1 m/sec^2.

Acceleration due to gravity is -9.8 m/sec^2, so the force has to supply an upward acceleration of 8.8 m/sec^2.

The force will be 8.8 times the mass.

Answer 3

I guess we ignore gravity (though the question does state "upward" that implies gravity of earth).

Force = Mass x Acceleration
Acceleration = Change in Velocity / Time

We know:
Time = 10s
Initial velocity = 10M/s
Final velocity = 0M/s
Velocity change = Final V - Initial V
= 0M/s - 10M/s
= -10M/s

SO...

F = Mass x (-10M/s / 10s)
F = Mass x (-1M per second squared)
F = Mass of rock(Kg) x -1M/(s squared)

The force is the negative of the force that launched the rock over 10s

The units will be in KgM/(s squared)

Answer 4

let mass of rock is m,
at the highest point, final velocity,v is zero, n initial velocity,u=10
Force is the rate of change of momentum right,
so F= m(v-u)/t, t is time taken=10

Answer 5

I have to think this through myself:

F = ma

a = dv/dt ------------> a*dt = dv --------------> v(t) = a*t + v(o)

We are given v(o) = 10m/sec and the time is 10 sec, the final velocity at v(10) = o, therefore:

0 = a*10sec + 10m/sec ---------------> a = -1m/sec^2

Since F = ma, we find that F = m(-1m/sec^2)