Student Problem 2: Launched up the Ramp The ramp is angled 3 degrees. The cart i

Question

Student Problem 2: Launched up the Ramp The ramp is angled 3 degrees. The cart is launched from the bottom of the track and travels 140 cm along the track, where it reaches it's highest point. Along the way up, it passed through a photogate at the 105 cm location. photogate 145 cm mass 250 Cm Cm How much kinetic energy did the cart start with at the bottom of the ramp? What speed does that correspond to? How much kinetic energy did the cart have when it passed through the photogate? What speed should this correspond to? 42

Explanation / Answer

at the bottom of the ramp,

a)

by using conservation of energy,

K.E=P.E

K.E=m*g*h

=m*g*d*sin(theta)

=(250*10^-3)*9.8*(140*10^-2)*sin(3)

=0.1795 J

b)

K.E=1/2*m*v^2

0.1795=1/2*(250*10^-3)*v^2

===> v=1.198 m/sec

at the place of photogate,

c)

by using conservation energy,

T.E=P.E + K.E

0.1795=m*g*h+K.E

0.1795=m*g*d*sin(theta) + K.E

0.1795=(250*10^-3)*9.8*(100*10^-2)*sin(3) + K.E

==> K.E=0.0513 J

d)

K.E=1/2*m*v^2

0.0513=1/2*(250*10^-3)*v^2

==> speed at photogate is, v=0.64 m/sec

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