A 60 kg skateboarder accelerates themselves at 1.5m/s2. How much force was required to do this?

Answers

Answer 1

Answer:

90 N

Explanation:

m = 60 kg

a = 1.5 m/s^2

F = m × a = 1.5 × 60 = 90 N


Related Questions

What is the net force acting on a 52 kg object that has a velocity of 8.0 m/s and is moving in a circle of radius 1.6 m? a. 4000N b. 20880N c. 2500N d. 3500N

Answers

Σf = m a
Σf = m v^2 / r
Σf = 52 8^2 / 1.6
Σf = 2080 N
The net force acting of the object will be "2080 N".

Given values are:

Mass, m = 52 kgVelocity, v = 8.0 m/sRadius, r = 1.6 m

As we know the formula,

→ [tex]\Sigma f = ma[/tex]

or,

→ [tex]\Sigma f = \frac{mv^2}{r}[/tex]

By putting the values, we get

         [tex]= \frac{52\times 8^2}{1.6}[/tex]

         [tex]= \frac{52\times 64}{1.6}[/tex]

         [tex]= 2080 \ N[/tex]

Thus the response above is appropriate.

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