Question

2013

1. which object has the greatest inertia?

(1) a 0.010 - kg bullet traveling at 90. m/s
(2) a 30. - kg child traveling at 10. m/s on her bike
(3) a 490 - kg elephant walking with a speed of 1.0 m/s
(4) a 1500 - kg car at rest in a parking lot

1. an 8.0 - newton wooden block slides across a horizontal wooden floor at constant velocity. what is the magnitude of the force of kinetic friction between the block and the floor?

(1) 2.4 n (2) 3.4 n (3) 8.0 n (4) 27 n

Explanation:

Step1: Analyze the motion state

Since the wooden block slides at a constant velocity, according to Newton's first law, the net force acting on it is zero. In the horizontal direction, the force of kinetic friction \(F_f\) and the applied force (assuming no other horizontal forces except friction, and if it's moving at constant velocity, the force causing motion is balanced by friction. If we assume the only horizontal force related to its motion - friction, and from the balance of forces in horizontal direction (as \(F_{net}=ma\), \(a = 0\) for constant velocity, so \(F_{applied}-F_f=0\), here if we assume the force related to its motion - if we consider the fact that for constant - velocity motion, the frictional force balances the "driving" force. But if we just consider the force balance in horizontal direction (no other horizontal forces), \(F_f\) is such that \(F_{net}=0\). In the vertical direction, \(F_N = G\) (normal force \(F_N\) equals the weight \(G\) of the block, \(G = 8.0N\)). And in horizontal direction, for constant velocity \(v\) (acceleration \(a=0\)), using \(F_{net}=F_{applied}-F_f = ma\), when \(a = 0\), if we assume the "driving" force is balanced by friction. But if there is no information about an applied force other than the fact that it's moving at constant velocity (which implies \(F_f\) is such that the net force is zero. If we assume the only horizontal force relevant to its motion - friction, then \(F_f\) must be such that there is no acceleration. In the most basic sense, for an object moving at constant velocity on a horizontal surface (no other horizontal forces), the force of kinetic friction is zero? No, wait, no - actually, if it's moving at a constant velocity, and assuming a force was applied to start it moving and then to keep it moving (but the problem doesn't mention an applied force. Wait, no - the key is Newton's first law. The sum of forces in the horizontal direction is zero. If we assume that the only horizontal force acting on the block (relevant to its motion) is the force of kinetic friction (which is a resistive force). But for it to move at a constant velocity, there must be a force equal in magnitude and opposite in direction to the force of kinetic friction. But if the problem is only asking for the magnitude of the force of kinetic friction, and we know from the vertical force balance \(F_N=G = 8.0N\) (since \(F_N - G=ma_y\), \(a_y = 0\)), and if we assume that in the horizontal direction, if it's moving at constant velocity \(F_{applied}=F_f\). But if the problem is missing some information (like an applied force), but wait - no, wait, another approach: using the formula \(F_f=\mu_kF_N\). But we also know from the motion (constant velocity \(a = 0\)), so \(F_{net}=F_{applied}-F_f=0\). If we assume that the "applied" force (the force that makes it move) is balanced by friction. But if we have no information about the applied force, but wait - no, wait, the problem might have a typo. Wait, no - wait, another way: if an object is moving at a constant velocity, the net force is zero. If we assume that the only horizontal force is the force of kinetic friction (which is impossible because then it would decelerate). So there must be an applied force \(F\) such that \(F - F_f=0\). But the problem doesn't mention \(F\). Wait, no - wait, looking back at the problem: it says "An 8.0 - newton wooden block slides across a horizontal wooden floor at constant velocity". The 8.0 N is the weight \(G\) of the block (\(G=mg = 8.0N\)), so \(F_N=G = 8.0N\) (vertical force balance). For horizontal motion, \(F_{net}=F_{applied}-F_f=ma\). Since \(a = 0\), \(F_{applied}=F_f\). But if we assume that the problem is using the fact that when moving at constant velocity, the frictional force is equal to the "driving" force. But if there is no information about the driving force, but wait - no, wait, maybe it's a misprint. Wait, no - another thought: if the block is moving at a constant velocity, the force of kinetic friction is equal to the force that is making it move. But if we assume that the only force related to its motion (in the sense of the problem's options) - wait, no, looking at the options: 2.4 N, 3.4 N, 8.0 N, 27 N. If we use \(F_f=\mu_kF_N\). For wood - on - wood, \(\mu_k\approx0.3\). Then \(F_f=\mu_kF_N\), \(F_N = G=8.0N\), \(F_f=0.3\times8.0N = 2.4N\)

Answer:

(1) \(2.4N\)