Megans mass is 28kg. She climbs the 4.8m ladder of a slide andreaches a velocity
ID: 1662863 • Letter: M
Question
Megans mass is 28kg. She climbs the 4.8m ladder of a slide andreaches a velocity of 3.2m/s at the bottom. How much work was doneby friction on megan? K=1/2(28)(3.2)2 = 143.36 J V= 4.8.(9.8) = 47.04 K=1/2(28)(47.04)2 =30978.67J W=K W= 30.8 KJ of work was done by friction on megan. Megans mass is 28kg. She climbs the 4.8m ladder of a slide andreaches a velocity of 3.2m/s at the bottom. How much work was doneby friction on megan? K=1/2(28)(3.2)2 = 143.36 J V= 4.8.(9.8) = 47.04 K=1/2(28)(47.04)2 =30978.67J W=K W= 30.8 KJ of work was done by friction on megan.Explanation / Answer
Be careful; friction always acts in the direction opposite theobject's motion, and hence is antiparallel to thedisplacement. Therefore, the work done by friction is alwaysnegative. Now, work done by dissipative forces is just thechange in mechanical energy Wdiss = Emech = Ef -E0 The initial mechanical energy is just the gravitational potentialenergy of Megan E0 = mgh and the final energy is just her kinetic energy at the bottom ofthe slide: Ef = (1/2)mvf2 so W = (1/2)mvf2 - mgh Inserting the given values vf = 3.2 m/s, h = 4.8 m, m =28 kg, and g = 9.8 m/s2, we have W = (1/2)(28 kg)(3.2 m/s)2 - (28 kg)(9.8m/s2)(4.8 m) = -1174 J = -1.174 kJ