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3 years ago

8

A block of mass 1.5 kg slides down an inclined plane has no friction and the black starts at height of 3 m, how much kinetic ene

rgy does the black have when it reaches the bottom

2 answers:

jok3333 [9.3K]3 years ago

7 0

Energy conservation
mgh = 1/2mv^2
1.5*9.8*3 = 1/2*1.5*v^2
44.1*2 = 1.5*v^2
88.2/1.5 = v^2
58.8 = v^2
v = 7.668 m/s

Sergio [31]3 years ago

5 0

<u>Answer:</u> The kinetic energy of the block is 44.1 J

<u>Explanation:</u>

As, the block is at some height, it will posses some potential energy.

When the block is released from the inclined plane, the block will start gaining kinetic energy as energy cannot be created nor be destroyed, explained by law of conservation of energy.

So,

Potential energy of the block = Kinetic energy of the block

Potential energy of the block is calculated by using the formula:

$E_p=mgh$

where,

$E_p$ = potential energy of the rock = ?

m = mass of the object = 1.5 kg

g = acceleration due to gravity = $9.8m/s^2$

h = height of the plane from where the block is thrown = 3 m

Putting values in above equation, we get:

$E_p=1.5kg\times 9.8m/s^2\times 3m\\\\E_p=44.1J=E_k$

Hence, the kinetic energy of the block is 44.1 J

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Answer:

a. FALSE

b.TRUE

C. FALSE

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hence we determine the distances between all the points

a.P(3,2,-4), Q(1,0,-4), R(2,1,1)

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For point PR

we have

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B. For point RP

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$RQ=\sqrt{(1-2)^{2} +(0-1)^{2} +(-4-1)^{2}}\\RQ=\sqrt{1+1+25}\\ RQ=\sqrt{27}$

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C.

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<u>To understand it better:</u>

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