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For example, if a box of 1.5 kg is subject to 5 forces which make it accelerate 2.0 m/s2 north-west, then the resultant force is directed north-west and has the magnitude equal to 1.5 kg × 2.0 m/s2 = 3.0 N. Often, however, we know the forces that act on an object and we need to find the resultant force.Export sql table to excel sheet
(a) Draw the free-body diagram for the block. (b) What is the net force acting on the block? (16.8 N) (c) What is the acceleration of the block? (3.35 m/s2) (d) How far up the incline does the block slide before coming to rest? (3.73 m) A 2.5-kg block slides down a 25( inclined plane with constant acceleration. The block starts from rest at the ...

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A 2.0 kg breadbox on a frictionless incline of angle is connected, by a cord that runs over a pulley, to a light spring of spring constant , as shown in Fig. 8-42. The box is released from rest when the spring is unstretched. Physics. A box is sliding down an incline tilted at an angle of 2.14° above horizontal.

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The block comes to rest at point C , a distance of 1.0 m along the table surface. How much has the internal energy of the system (block + table) increased? Example 14.1 Dissipation of Energy by Friction A block of mass 10.0 kg starts at point A at a height of 2.0 m above the horizontal and slides down a frictionless incline (Fig. 14.1).

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A 200-g block is released from rest at a height of 25 cm on a frictionless 30 ∘ incline. It slides down the incline and then along a frictionless surface until it collides elastically with an 800-g block at rest 1.4 m from the bottom of the incline.

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6. A 3.0-kg block slides on a frictionless 20° inclined plane. A force of 16 N acting parallel to the incline and up the incline is applied to the block. What is the acceleration of the block? a. 22.0 m/s down the incline b. 25.3 m/s up the incline c. 22.0 m/s up the incline d. 23.9 m/s down the incline e. 3.9 m/s2 up the incline 7.

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A 4.00 kg block rests on a 30.0 degree incline as shown in the figure. If the coefficient of static friction between the block and the incline is 0.700, with what magnitude a horizontal force must ...

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Oct 13, 2009 · (1) 1.0 kg (3) 0.50 kg (2) 2.0 kg (4) 1.5 kg The diagram below shows a sled and rider sliding down a snow-covered hill that makes an angle of 30.° with the horizontal. Which vector best represents the direction of the normal force, FN, exerted by the hill The diagram below represents a block sliding down an incline.

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Find the torque of the normal force acting on the block about its centre. Let R = Since R and mg cos θ pass through the centre of the cube, there will be no torque due to R and mg cos θ. The only torque will be produced by mg sin θ. i = F × r (r = a/2) (a = ages of the cube) i = mg sin θ × a/2 = 1/2 mg a sin...

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A 4 O—kg block is pushed up a 360 incline by a force of magnitude F applied parallel to the incline. When F is 31 N, it is observed that the block moves up the incline with a constant speed. What value of F would be required to lower the block down the incline at constant speed? b. c. 27 15 13 17 19 N N N N