A 2.00 kg box is moving to the right with a speed of 9.00 m\/s, on a horizontal frictionless surface. At t = 0, a horizontal force is applied to the box. The force is directed to the left and has a magnitude F(t) = 6.00 N A)<\/p>\n
A) What distance does the box move from its position at t = 0 before its speed is reduced to zero?<\/p>\n
B) If the force continues to be applied what is the speed of the box at t = 3.00s.<\/p>\n
Answer:-\u00a0<\/strong>We know that the force is applied in the opposite direction of the box’s motion, so it will cause the box to decelerate. We can use Newton’s second law to find the acceleration of the box:<\/p>\n F = ma<\/p>\n 6.00 N = 2.00 kg * a<\/p>\n a = 3.00 m\/s^2<\/p>\n Now that we know the acceleration, we can use the equation for the distance traveled by an object under constant acceleration:<\/p>\n x = v_0t + (1\/2)at^2<\/p>\n x = 9.00 m\/s * t + (1\/2) * 3.00 m\/s^2 * t^2<\/p>\n We can solve for t by setting the velocity to zero:<\/p>\n 0 = 9.00 m\/s + (1\/2) * 3.00 m\/s^2 * t^2<\/p>\n 0 = (1\/2) * 3.00 m\/s^2 * t^2<\/p>\n t^2 = 6.00 s^2<\/p>\n t = 3.00 s<\/p>\n Now that we know the time it takes for the box to stop, we can find the distance it travels by plugging t = 3.00 s into the equation for x:<\/p>\n x = 9.00 m\/s * 3.00 s + (1\/2) * 3.00 m\/s^2 * 3.00 s^2<\/p>\n x = 27.00 m<\/p>\n Therefore, the box will travel 27.00 m before its speed is reduced to zero.<\/p>\n","protected":false},"excerpt":{"rendered":" A 2.00 kg box is moving to the right with a speed of 9.00 m\/s, on a horizontal frictionless surface. At t = 0, a […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[27],"tags":[],"yoast_head":"\n