An exernal force F=150N making an angle 53° with the horizontal is applied to a block if mass 10kg
.find the displacement of the block in four seconds .initial velocity of the block is 10 m/s towards right.
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F =150 N
Since the the block displaces horizontally, we need to take the horizontal (cos) component of the Force.
F= 150 cos 53 = 150*3/5=90 N
F= ma
90 = 10*a
a= 9m/s²
S= ut + 1/2 at²
S= 10*4 + 1/2*9*4²
S= 122m
Since the the block displaces horizontally, we need to take the horizontal (cos) component of the Force.
F= 150 cos 53 = 150*3/5=90 N
F= ma
90 = 10*a
a= 9m/s²
S= ut + 1/2 at²
S= 10*4 + 1/2*9*4²
S= 122m
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