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suppose that the stone is launched with a speed of 3 m/s and travels 40 m before coming to rest. what is the approximate magnitude of the friction force on the stone? 0 n 2 n 20 n 200 n



Answer :

The approximate magnitude of the friction force of the stone is 2N.

Given values:

Mass(m)= 20 Kg

Velocity(v)= 3m/s

Distance(x)= 40m

Kinetic energy is given by the expression:

[tex]Ek=1/2mV^{2}\\ Ek=1/2*20Kg*(3m/s)^{2} \\Ek=90J[/tex]

If we divide the kinetic energy by the distance traveled we get the friction force:

[tex]Ff=Ek/x\\Ff=90J/40m\\Ff=2.25N[/tex]

Approximate magnitude of the friction force is: [tex]Ff=2N[/tex]

Result is 2N.

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