Question #3611

A force is applied on a body ,mass 50 kg, for 8 sec. After that the body cover 80m distance in 5 sec in coming to rest.Find the force applied.

Expert's answer

Index “1” means first part of path and index “2” the second part

V is the velocity at time 8sec

F_{f} is a friction force

F_{1} = F-F_{f} = ma_{1}

a_{1} = (F-F_{f})/m

V = a_{1} t_{2}

F_{2} = F_{f} = ma_{2}

a_{2} = F_{2}/m

V = a_{2} t_{2} = F_{2}/m t_{2} (*)

a_{1} t_{1} = a_{2} t_{2}

(F-F_{f})/m t_{1} = F_{f}/m t_{2}

F = F_{f} (1+t_{2}/t_{1} ) (**)

A_{2} = F_{f} S_{2}

E_{1} = (mV^{2})/2

A_{2} = E_{1}

Let put here the V from (*)

F_{f} S_{2} = m 1/2 F_{f}^{2}/m^{2} t_{2}^{2}

so,

F_{f} = (2mS_{2})/(t_{2}^{2} )

(**):

F = (2mS_{2})/(t_{2}^{2} ) (1+t_{1}/t_{2} ) = (2*50*80)/5^{2} (1+8/5) = ** 832 N**

V is the velocity at time 8sec

F

F

a

V = a

F

a

V = a

a

(F-F

F = F

A

E

A

Let put here the V from (*)

F

so,

F

(**):

F = (2mS

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