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(1) It can be seen from the image that the acceleration of the falling block is a = △ V △ T = 2? 0 1.5? 0.5 = 2m/S2;

(2) Using the area method of v-t image, the upward sliding displacement is calculated.

S= 12×4×0.5m= 1m?

That is, the maximum sliding distance of the block is1m. ..

(3) Let the mass of the block be m and the sliding friction coefficient between the block and the inclined plane be μ,

When an object slides upward along an inclined plane, the force is as shown in the figure.

According to Newton's second law

mgsinθ+μmgcosθ=ma 1

a 1=△v△t=0? 40.5=? 8m/s2

When an object slides down an inclined plane, the force is shown in the figure.

According to Newton's second law

mgsinθ-μmgcosθ=ma2

a2=△v△t=2? 0 1.5? 0.5 = 2m/S2;

From the above two formulas, it can be solved by substituting data.

θ=30

That is, the inclination of the inclined plane is 30.

Answer: (1) The falling acceleration is 2m/s2; ?

(2) The maximum sliding distance of the block is1m; ;

(3) The inclination of the inclined plane is 30.