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The car in the figure, M = 19kg is subjected to an F = 100N horizontal constant. A block m = 1kg, slides on a fixed inclined plane, with Θ = 37° its angle to the horizontal and g = 10m /s^2. Calculate the acceleration of the block in relation to the car. Despise friction.
The answer is: 10,33m/s^2

The car in the figure M 19kg is subjected to an F 100N horizontal constant A block m 1kg slides on a fixed inclined plane with Θ 37 its angle to the horizontal class=

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Answer:

you can take the car an inertial frame after introducing pseudoforce acting on block which is equal to product of

acc. of car (100/19) and mass of block( m kg) i.e. m100/19

this pseudo force will act on mass towards left

In addition to this there is gravitational force acting on mass in vertically downward direction.

Explanation:

so

net force acting on block along the surface of inclined plane is= m (100/19)cos37 + m 10 sin 37

so net acc. down the plane

acc= (100/19)Cos 37 + 10 sin 37

= 4.21 + 6.12

= 10.33 m/s^2

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