2. Refer to the diagrame here (a side view, not to scale). A block (m=6.25kg) is

Question

2. Refer to the diagrame here (a side view, not to scale). A block (m=6.25kg) is sliding to the right on this level, frictionless surface. At point A, its velocity of 1.74 m/s. A force fx, always directed in the +x direction, is exerted on the block while it is sliding in the range of 0 < x < 10.0m. A graph of Fx vs. x is shown below. The block then slides up the frictionless 30.0 slope for 1.00m and then becomes a projectile, then lands back at the original level (the x-axis). Assume g=9.8m/s^2 and ignore all air drag.

Find the mechanical power being delivered by the force of gravity to the block during its final moment as a projectile.

2. Refer to the diagram here (a side view, not to scale). A block (m 6.25 kg) is sliding to the right on this level, frictionless surface. At point A, its velocity of 1.74 m/s. A force Fx, always directed in the +x-direction, is exerted on the block while it is sliding in the range 0 sxs 10.0 m. Agraph of Fx vs. x is shown below. The block then slides up the frictionless 30.0° slope for 1.00 m and then becomes a projectile, then lands back at the original level (the x-axis). Assume g = 9.80 m/s? and ignore all air drag Find the mechanical power being delivered by the force of gravity to the block during its final moment as a projectile. IIm 30° x (m) 0 2 4 6 8 10 F, (N) 3.0 '20 40 6.0 8.0 10.0*

Explanation / Answer

mass of block = m=6.25 Kg

Va = 1.74 m/s

At A block has KE = 0.5 X 6.25 X 1.74M/s^2 = 9.46 J

By applying a force F(x) for distance of 10 m additional energy is imaparted to the block , the amount of energy imparted is the area under the curve F(x) Vs x = 33 J

Total Energy when it reaches ramp = 33+ 9.46 J = 42.46 J

When it clims ramp for 1m distance it climbs the ramp by a vertical height = 1 sin 30 = 0.5m

So PE gained on ramp = 0.5 m X 9.8 X 6.25 Kg = 30.63 J

KE at top of ramp = 42.46 J - 30.63 J = 11.83 J

11.83 = 0.5 X 6.25 X Vf^2

Vf= sqrt {11.83 /(0.5 X 6.25)} = 1.95 m/s

Work done by gravity in the projectile motion = Mgh = 6.25 Kg X 9.8 m/s^2 X 0.5 m = 30.63 J

The projectile travel 0.5 m from ramp top to ground level ,let us say in time t .

The projectile has initial velocity in vertical upward direction of 1.95 X sin 30 = 0.975 m/s

0 = 0.5 + 0.975 t - 0.5 X 9.8 X t^2

solving above quadratic equation we get t = 0.434 seconds

Power =30.63 J/0.434 = 70.6 watts

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