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3 years ago

How much force is needed to accelerate a 3 kg skateboard at 5 m/s2

Physics

2 answers:

TEA [102]3 years ago

4 0

Answer:15 newton force is required.

Explanation:

We know that

Force equal to mass x acceleration

3 kg x5 m/ s square

15( kg x m/s square)

15( newton)

15 N

Umnica [9.8K]3 years ago

3 0

The problem is fairly simple if you use the formula.

Formula: Force = (mass) x (acceleration)

3 x 5 = 15 newtons

Note: 3 kg is the mass of the skateboard, and 5 m/s2 is the acceleration.

Answer: 15 newtons

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A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas

Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

$V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s$

$a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2$

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3 years ago

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3 years ago

As a box slides down a ramp, friction does 23.0 joules of work. At the bottom of the ramp, the box has 3.8 joules of kinetic ene

tensa zangetsu [6.8K]

Answer:

The high of the ramp is 2.81[m]

Explanation:

This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.

If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.

We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.

$E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]$

Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.

$E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\$

And therefore

$m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]$

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2 years ago