Question #246335
A 10 kg ball falls form the height of 100 m (a) calculate the vertical speed of the ball during the first 4 seconds ,(b) calculate the height of the ball above the gorund during the first 4 seconds ,(c) calculate the kinetic and potential energies of the ball.(d) determine the total mechanical energy of the ball?
1
Expert's answer
2021-10-05T13:37:02-0400

Let positive y-axis be pointing up. Then


v(t)=v0gt=gt,v0=0v(t)=v_0-gt=-gt, v_0=0

y(t)=y0+v0tgt22=y0gt22,v0=0,y0=100 my(t)=y_0+v_0t-\dfrac{gt^2}{2}=y_0-\dfrac{gt^2}{2}, v_0=0, y_0=100\ m

(a)


v(4)=9.81 m/s2(4 s)=39.24 m/sv(4)=-9.81\ m/s^2(4\ s)=-39.24\ m/s

(b)


y(4)=100 m9.81 m/s2(4 s)22=21.52 my(4)=100\ m-\dfrac{9.81\ m/s^2(4\ s)^2}{2}=21.52\ m

(c)


P=mgyP=mgy

Pt=0=10 kg(9.81 m/s2)(100 m)=9810 JP|_{t=0}=10\ kg(9.81\ m/s^2)(100\ m)=9810\ J

Pt=4=10 kg(9.81 m/s2)(21.52 m)=2111.112 JP|_{t=4}=10\ kg(9.81\ m/s^2)(21.52\ m)=2111.112\ J

K=mv22K=\dfrac{mv^2}{2}

Kt=0=0K|_{t=0}=0

Kt=4=10 kg(39.24 m/s)22=7698.888 JK|_{t=4}=\dfrac{10\ kg(-39.24\ m/s)^2}{2}=7698.888\ J

(d)


E=P+K=constE=P+K=const

E=9810 J+0 J=9810 JE=9810\ J+0\ J=9810\ J

=2111.112 J+7698.888 J=2111.112\ J+7698.888\ J

The total mechanical energy of the ball is 9810 J and is conserved during the motion.



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