Explanation Given data: An ice cube weighing 20 g is pushed horizontally with a speed of 0.8 m/s from the top of the incline, as shown in the figure. Formula used: Write the expression for displacement (motion under a constant acceleration). s = u t + 1 2 a t 2 Here, u is the initial velocity, s is the distance travelled, t is the time (in seconds) and a is the acceleration. Write the expression for the Pythagoras theorem. h 2 = b 2 + p 2 Here, h is the hypotenuse of the triangle, b is its base and p is its perpendicular. Write the expression for velocity (motion under a constant acceleration). v = u + a t Here, u is the initial velocity, t is the time (in seconds) and a is the acceleration. Explanation: Consider a coordinate axis with the origin at the starting point of the motion, considering the direction down the incline to be the positive y-axis and that along the initial velocity to be the positive x-axis. Let θ be the inclination. Take the velocity component along the y-axis as v y , that along the x-axis as v x , the acceleration along the y-axis as a y and that along the x-axis as a x . As gravity is acting on the ice cube, resolving gravity down the incline. Therefore, a y equals to g sin θ . Calculate the slope ( θ ) . From the given figure: tan θ = 20 cm 30 cm tan θ = 2 3 θ = tan − 1 2 3 = 33.69 ° Now, calculate the distance travelled by the ice down the incline. Consider a triangle with the base ( b ) equal to 30 cm and the perpendicular ( p ) equal to 20 cm . Then, the distance down the incline will be the hypotenuse ( h ) . Refer to the expression for the Pythagoras theorem. h 2 = b 2 + p 2 Substitute 30 cm for b and 20 cm for p . h 2 = ( 30 cm ) 2 + ( 20 cm ) 2 h = ( 30 cm ) 2 + ( 20 cm ) 2 h = 36.055 cm Calculate the time taken by the ice cube to reach the bottom of the incline. From the given figure, v y is zero and a y is g sin θ . Refer to the expression for motion under constant acceleration to calculate the displacement. s = u t + 1 2 a t 2 Substitute v y for u , h for s and a y for a . h = ( v y ) t + 1 2 ( a y ) t 2 Substitute 0 m/s for v y , 36.055 cm for h and g sin θ for a y . [ 36.055 cm ( 1 m 100 cm ) ] = ( 0 m/s ) t + 1 2 ( g sin θ ) t 2 Substitute 9

2nd Edition

ISBN: 9780134601823

Author: ETKINA, Eugenia, Planinšič, G. (gorazd), Van Heuvelen, Alan

Publisher: Pearson,

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Chapter 7, Problem 33P

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The speed of the ice cube when it reaches the bottom of the incline, given that an ice cube weighing 20 g is pushed horizontally with a speed of 0.8 m/s from the top of the incline, as shown in the figure.

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Chapter 7, Problem 32P

Chapter 7, Problem 34P

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