Two forces act on the hockey puck of mass of 030 kg as it...

Two forces act on the hockey puck of mass of 0.30 kg as it slides on the horizontal, frictionless surface of an ice rink. The forces magnitudes and directions are described in the figure 1 below. Determine both the magnitude and the direction of the puck’s acceleration. Figure 1: Hockey puck moving on a frictionless surface. F1 = 5.0 N 20 degrees F2= 8.0 N 60 degrees

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00:01 Hello, the question is taken from trigonometry and the question is two forces acting on the hokey book of mass 0 .3 kg.

00:09 So this is 0 .3 kg mass as it slide on the horizontal frictionless surface of an ice rink.

00:17 The force of magnitude and direction are described in figure below.

00:21 Determine both the magnitude and direction of book acceleration.

00:25 Hokie book moving on a frictionless surface.

00:28 F1 is 5 newton.

00:29 So given that 5 newton is acting at an angle 20 degree and other force is 8 newton so acting at an angle 60 degree so this is 5 newton mass of it is 0 .3 kg okay so we first need to evaluate the acceleration of the book we know that let us evaluate the first net force there are two forces here one is f y so this will be 8 sign 6 degree and downward component is 5 sine 20 degree horizontal component are 8 460 degree plus 5 course 20 degree so force alone x will be equal to 8 course 60 degree plus 5 goes 20 degree so and f y vertical component is 8 sine 60 degree plus minus sensitive downward direction so it will be minus 5 sign 20 degree so from here fx will be equal to let me evaluate the value of fx 8 by 2 that is 4 plus 5 course 20 so that will be 8 .698 mean newton and that will be 8 sine 60 degree minus 5 sign 20 so 20 degree so that will be 5 .22 okay so this is the required amount of force that is acting on the book along the x and y direction now we need to evaluate the net acceleration acceleration is net force which is i'm taking the magnitude initially f y square plus f x squared divided by mass so that is that will be squared 5 .22 square plus 8 .698 square divided by mass is 0 .3...

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