A 1 kg ball rolls off a 32 m high cliff, and lands 27 m from the base of the cli

Question

A 1 kg ball rolls off a 32 m high cliff, and lands 27 m from the base of the cliff. Express the displacement and the gravitational force in terms of vectors and calculate the work done by the gravitational force. Note that the gravitational force is < 0, -mg, 0 >, where g is a positive number (+9.8 N/kg). (Let the origin be at the base of the cliff, with the +xdirection towards where the ball lands, and the +y direction taken to be upwards.)

Work ? W =  J

Explanation / Answer

m=1kg,

y1 =32m, at top of cliff; x1 =0

y2=0 when ball landed; x2=27m

=> displacment r =(27-0)i^+(y2-y1)j^ =(0-32)j^ =27i^-32j^ m

graviational force F =m*g =1*9.81 (-j^) =-9.81j^ (since gravity is always acts downward)

work W=(Fxi^+Fyj^) . r (It is a dot product& Fx=0 as there is no horizontal force, Fy =m*g =9.81)

=>W=(0i^-9.81j^).(27i^-32j^)

=>W=-9.81*-32 =313.92J (j^.j^ =1 & j^.i^=0)

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