To get off a frozen lake, a 70 kg person removes his shoe of mass 0.175 kg and throws it horizontally away from the shore at a velocity of 3.2 m/s. if the person is 5.15 m from the shore, how long do they take to reach the shore?

the person will be in the shore at 10.73 minutes after launch the shoe.

Explanation:

For this we will use the law of the lineal momentum.

Also,

L = MV

where M is de mass and V the velocity.

replacing,

wher Mi y Vi are the initial mass and velocity, Mfp y Vfp are the final mass and velocity of the person and Mfz y Vfz are the final mass and velocity of the shoe.

so, we will take the direction where be launched the shoe as negative. then:

(70)(0) = (70-0.175)() + (0.175)(-3.2m/s)

solving for ,

=

= 0.008m/s

for know when the person will be in the shore we will use the rule of three as:

1 second -------------- 0.008m

t seconds-------------- 5.15m

solving for t,

t = 5.15m/0.008m

t = 643.75 seconds = 10.73 minutes

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

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