As the object is in motion hence kinetic friction will take place
kinetic friction = μk×N=1.1×21×10\mu_k\times N=1.1\times21\times10μk×N=1.1×21×10 =231N231 N231N
lets apply work energy theorem
∑W=ΔK\sum W=\Delta K∑W=ΔK
WN+Wf+Wmg+WF=ΔKW_N+W_f+W_{mg}+W_F=\Delta KWN+Wf+Wmg+WF=ΔK
0−(231×23)+0+WF=12×21×0.920-(231\times23)+0+W_F=\frac12\times21\times0.9^20−(231×23)+0+WF=21×21×0.92
WF=5321.5JW_F=5321.5 JWF=5321.5J
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