Question

To test a slide at an amusement park, a block of wood with mass 3.00 kgkg is released at the top of the slide and slides down to the horizontal section at the end, a vertical distance of 23.0 mm below the starting point. The block flies off the ramp in a horizontal direction and then lands on the ground after traveling through the air 30.0 mm horizontally and 40.0 mm downward. Neglect air resistance.
How much work does friction do on the block as it slides down the ramp?

Answer 1

Answer:

511.1 J

Explanation:

We are given that

Mass of wood block=m=3 kg

Vertical distance,h=23 m

Horizontal distance =x=30 m

Distance traveled in downward direction y=40 m

Initial velocity,u=0

$y=ut+\frac{1}{2} gt^2$

Where $g=9.8 m/s^2$

$40=0+\frac{1}{2}(9.8)t^2=4.9t^2$

$t^2=\frac{40}{4.9}$

$t=\sqrt{\frac{40}{4.9}}=2.86 s$

$x=v_x\times t$

$30=v_x(2.86)$

$v=v_x=\frac{30}{2.86}=10.49 m/s$

By work energy theorem

Change in kinetic energy=Work done= mgh-W

$\frac{1}{2}mv^2-\frac{1}{2}mu^2=3\times 9.8\times 23-W$

$\frac{1}{2}(3)(10.49)^2-0=676.2-W$

$W=676.2-\frac{1}{2}(3)(10.49)^2=511.1 J$

Hence, the work done due to friction on the block as it slides down the ramp=511.1 J