Physics, 26.07.2019 11:00 Amazingpandakid

Ablock of mass 10 kg moves from position a to position b shown in the figure above. the speed of the block is 10 m/s at a and 4.0 m/s at b. the work done by friction on the block as it moves from a to b is most nearly

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We have that the block is moving horizontally. Hence, its potential energy due to gravity stays the same. The only change in its mechanical energy is the one due to the change of speed. This reduction of its kinetic energy, due to the conservation of energy, is equal to the work that friction does. We have that at A the kinetic energy is : K=1/2*m*u^2=10*10*10/2=500J. At B, we have that K=1/2*10*16=80J. Sine we have that the initial value is 500, the work from the friction force (opposite to the movement of the object) is 80-500=420J.

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Is a vector quantity that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.

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initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

$a = 32 ft/s^2$

so as we know that

$a = \frac{v^2}{r}$

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

$a_c = 32 ft/s^2$

$a_t = 0$

now radius of curvature is given as

$r = \frac{v^2}{a_c}$

$r = \frac{150^2}{32}$

$r = 703.125ft$

so the radius of curvature just after it is dropped is 703.125 ft

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Ablock of mass 10 kg moves from position a to position b shown in the figure above. the speed of the...

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