Un trineo de 8.00kg se mueve en línea recta sobre una superficie horizontal sin fricción. En cierto punto su rapidez es de 4.00m/s 2.50m más adelante es de 6.00m/s

An 8 kg sled mass moves in a straight line on a surface without horizontal friction at a certain point its speed is 4 m / s 2.50 m later its speed is 6 m / s using the work and energy theorem, determine the force acting on the sled assuming it is constant and acts in the direction of the sled movement

The force acting on the 8 kg sled has a value of 32 N.

Explanation:

We have initially look for the work that is applied to the sled, and this will be nothing more than the difference in kinetic energy from the end point to the initial, then:

ΔEc = Ec₂ - Ec₁

ΔEc = 0.5 * m * [V₂² - V₁²]

ΔEc = 0.5* 8 kg * [(6m / s) ² - (4m / s) ²]

ΔEc = 80 J

So, the work is 80 J, now we know that work is a relation distance and force , such that:

W = F * d

F = W / d

F = (80 J) / (2.50 m)

F = 32 N

Therefore, the force acting on the sled is 32 N.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)