# Answer to Question #3611 in Mechanics | Relativity for Vikash

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

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

V is the velocity at time 8sec

F

_{f}is a friction forceF

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

_{1}= (F-F_{f})/mV = a

_{1}t_{2}F

_{2}= F_{f}= ma_{2}a

_{2}= F_{2}/mV = 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})/2A

_{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**
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