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4 years ago

12

determine the quantity of work done when a crane lifts a 100-n block from 2m above the ground to 6m above the ground

2 answers:

Mekhanik [1.2K]4 years ago

6 0

100n divided by 8m because 2m and 6m are like terms. Hope that helps:)

xz_007 [3.2K]4 years ago

3 0

Work done = Force.Δx
Work done = 100(6-2). cos(180°) = -100 (4) = -400 J.
Work done is negative because force acting on the block and its displacement are in opposite direction. Hope this helps!

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Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

So

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

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4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two

Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=mV_{o}+MU_{o}$ (2)

$p_{f}=(m+M)V_{f}$ (3)

$m=1200 kg$ is the mass of the first car

$V_{o}=20 m/s$ is the velocity of the first car, to the North

$M=1400 kg$ is the mass of the second car

$U_{o}=-22 m/s$ is the mass of the second car, to the South

$V_{f}$ is the final velocity of both cars after the collision

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg}$ (6)

Finally:

$V_{f}=-2.61 m/s$ (7) This is the resulting velocity of the wreckage, to the south

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