A 4.4 kg block is pushed 6.6 m up a rough 35 Solution (a) the initial kinetic en

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(a) the initial kinetic energy of the block -->this is what i got KE = 0.5*m*v0² (.5)(6)(2²)=12 J (b) the work done by the 75-N force; -->= 75*8.0*cos37º = 479.16 J (c) the work done by the friction force; -->Friction force = 25N so Wf = 8*25 = -200 J (d) the work done by gravity; -->mgh h=8(sin 37)=4.81 , mg=58.8 W=(58.8)(4.81)= -283 (e) the work done by the normal force; -->is no motion in the direction of the normal force, so the work = 0 work_Ke = Kei = (1 / 2) * m * v ^ 2 work_75 = 75 * 8 * Cos(A) work_friction = Fk * 8 work_g = m * g * d * Sin(A) With rounding, I get: a) Kei = 12 J b) work_75 = 479 J c) work_friction = -200 J d) work_g = -283 J e) work_N = 0 ---->Total_work = 8 J Conservation of energy Kef = (1 / 2) * m * vf ^ 2 = Total_work ---->vf = sqr(2*8/6) = 1.63 m/s

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