- #1

himetx

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A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track (as shown in the figure attached for#007). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.8 m/s2 .

m=0.405 kg

H=5.2 m

h=2 m

rough area=1.2m

μ=0.3

A. What is the speed v of the block when it leaves the track? Answer in units of m/s.

B. What is the horizontal distance x the block travels in the air? Answer in units of m.

C. What is the total speed of the block when it hits the ground? Answer in units of m/s.

The initial height of the block when it is at rest at the top of frictionless track = H

Height of the point where the block leaves the track horizontally = h

Change in height = H - h

Loss in gravitational potential energy of block=mg(H-h)

Gain in kinetic energy of block as it leaves the track =Loss in gravitational potential energy of block

Gain in kinetic energy =(1/2)mv^2

(1/2)mu^2=mg(H - h)

u = sq rt 2g (H - h)

If H and h are in meter then ball leaves the track with a speed = sq rt19.62 (H-h) m/s

Final kinetic energy of block (KE) =initial potential energy of block (PE)

(1/2)mV^2=mgH

V= sq rt 2gH

the speed of the block when it hits the ground is sq rt 2gH

If H is in meter then speed is [ sq rt 19.62 H ] m/s

m=0.405 kg

H=5.2 m

h=2 m

rough area=1.2m

μ=0.3

A. What is the speed v of the block when it leaves the track? Answer in units of m/s.

B. What is the horizontal distance x the block travels in the air? Answer in units of m.

C. What is the total speed of the block when it hits the ground? Answer in units of m/s.

The initial height of the block when it is at rest at the top of frictionless track = H

Height of the point where the block leaves the track horizontally = h

Change in height = H - h

Loss in gravitational potential energy of block=mg(H-h)

Gain in kinetic energy of block as it leaves the track =Loss in gravitational potential energy of block

Gain in kinetic energy =(1/2)mv^2

(1/2)mu^2=mg(H - h)

u = sq rt 2g (H - h)

**The ball leaves the track with a speed = sq rt 2g ( H - h)**__PART A.__If H and h are in meter then ball leaves the track with a speed = sq rt19.62 (H-h) m/s

__I understand how to get to this point, but I don't know how to account for the friction that needs to be considered to determine the velocity.__**If V is speed with which block hits the ground.**__PART C.__Final kinetic energy of block (KE) =initial potential energy of block (PE)

(1/2)mV^2=mgH

V= sq rt 2gH

the speed of the block when it hits the ground is sq rt 2gH

If H is in meter then speed is [ sq rt 19.62 H ] m/s